If zy < xy < 0, is |x - z | + |x| = |z|?

Hi, rgtiwari
Draw a number line and put 0 in the middle of it. now look at the data you are given. zy<xy<0.
we don't know if y is positive or negative. if y>0 than z is on the left of x which is on the left of zero => z<x and both z & x are originally negative.
the opposite if y<0, x will be on the right of zero and z will be on the right of x. => z>x. and both are originally positive. in both cases it seems that the equation is correct. but don't bother to check that just look at the extra data and continue from there.
If you find it hard to understand with variables just use numbers instead.
now let's look at (1) it tells you that z<x so we know that this is the first case only => the equality is definitely correct (just use numbers that maintain the data in the first case) - Sufficient.
(2) it tells you that y>0. again we are at the first case. - Sufficient.

pay attention that this is not a coincidence that the extra data given leads you in (1) and (2) to the same conclusion. if it doesn't you should suspect whether it's D or not.

Hope that helps.

Instead of plugging in numbers, use the conceptual understanding of mods and inequalities here.

Important points to remember:
1. If ab is negative, either a or b but not both should be negative.
2. If a is positive, |a| = a; if a is negative, |a| = -a
3. If |a| > |b| but a < b, a must be negative.

Given: zy<xy<0
This means zy and xy are both negative. This is possible only if [highlight]either y is negative (and z and x are positive)[/highlight][highlight][/highlight] OR [highlight]both z and x are negative but y is positive[/highlight]. Also, since zy is 'more negative' than xy, the absolute value of z is higher than that of x i.e. [highlight]|z| > |x|[/highlight]

Question: Is |x-z| = |z| - |x|?

Let's look at stmnt 2 first. y > 0
This means that both x and z are negative. We know that if z is negative, |z| = -z. Therefore, the question now is: [highlight]Is |x-z| = x - z[/highlight]?
Again, we know that |x-z| is equal to x-z if x-z is positive. So, is x-z positive? We know that x and z are both negative and z is more negative than x (e.g. x = -2 and z = -3). Therefore, x-z will be positive.
So the answer to Is |x-z| = x - z? is 'YES'
Sufficient.

Stmnt 1: z < x
We know from above that |z| > |x|. If still z < x, then z must be negative (which implies that both x and z must be negative). This is exactly same as the situation above in stmnt 2 hence this statement will be sufficient alone.

Answer D.

Since zy < 0 and xy < 0, both z and x have opposite sign to y, so they must be either both positive or both negative. In other words, we know that xz > 0.

(1) Given that z < x, when both z and x are negative, |z - x| + |x| = -z + x + (-x) = -z = |z| TRUE
z and x cannot be both positive, because then y would be negative, and from zy < xy we would obtain that z > x.
Sufficient.

(2) Knowing that y > 0, we can deduce that both z and x are negative. In addition, from zy < xy it follows that z < x, and we are in the same case as above.
Sufficient.

Answer D

If | z - x |= |z| - |x| assuming x & z have same sign
| x - z |= |z| - |x| ; Since | x - z | = | z - x |

Implies |z| - |x| = |x| - |z|
which is defn not true
z = 1 , x = 2
Plugging, we get
- 1 = 1 ?!?

Have I misunderstood something?

Before discussing this part, I have discussed in my post that absolute value of z must be greater than absolute value of x.
If absolute value of z is not greater than absolute value of x, then | z - x |= |z| - |x| does not hold when x and z have the same sign.

Since |z| must be greater than |x|,
| x - z | = |x| - |z| does not hold.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If zy < xy < 0, is | x - z | + |x| = |z|?

(1) z < x
(2) y > 0

We get y(z-x)<0, is |x-z|=|z|-|x|?, zx>0 and z>x? if we modify the question.
There are 3 variables (x,y,z) and 1 equation in the original condition,
2 more equations in the given conditions, so there is high chance (C) will be the answer.
But condition 1=condition 2 answering the question 'no' and sufficient.
The answer therefore becomes (D).

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

Hi Bunuel,

In step A, How did you get the below:
|x−z|=−x+z
Why did you take negative sign on removing the Modulus?

Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

So, to answer your question: we have that \(z>x\), which is the same as \(x-z<0\), thus \(|x-z|=-(x-z)=-x+z\).

Hope it's clear.

but for the equality to hold true done we need to know for sure that /z/ > /x/ and that they have the same sign??

Both points have been considered.

The first sentence of the solution states:
"zy < xy < 0 implies both 'zy' and 'xy' are negative and zy is more negative i.e. has greater absolute value as compared to xy. Since y will be equal in both, z will have a greater absolute value as compared to x."

Since |zy| > |xy|,
|z||y| > |x||y|
Divide by |y| on both sides
|z| > |x|

Also, note the other highlighted sentence:
"Now think, when will | z - x | = |z| - |x| ? It happens when x and z have the same sign."

why in case A, have you expanded |x-z| as -x+z, but |x| and |y| as x and y?? shouldnt |x| and |y| be expanded as -x and -y as well?

A. If \(y<0\) --> when reducing we should flip signs and we'll get: \(z>x>0\).

Since \(z>x\) --> x - z < 0, thus \(|x-z|=-(x-z) =-x+z\);
Since \(x>0\) and \(z>0\) --> \(|x|=x\) and \(|z|=z\).

HI Bunuel

I have a question. I answered the question correctly but I'm not sure if the steps I took are correct.

I looked at the equation |x-Z| + |x| = |z| and noticed that with the absolute, each term will be positive. So I squared each term.

I ended up with 2X^2 -2XZ = 0. With a little rearranging, I got X=Z. So the question is really asking if Z=X

From here I was able to answer both statements quickly.

ST1: Z<X Sufficient. Z doesn't equal X
ST2: Y>0 Sufficient. If Y is positive than Z and X are negative. The inequality lists ZY<ZX so Z doesn't equal X.

Question - Was squaring each term legit in this case because the expression has the same positive sign due to the absolute value sign?

Many thanks in advance

That's not correct. You cannot square individual terms the way you did. Formula of square of the sum of two numbers is \((a + b)^2 = a^2 + 2ab + b^2\). What I mean is that if you square \(a + b = c\) you get \((a + b)^2 = c^2\) NOT \(a^2 + b^2 = c^2\). So, if you square \(|x - z| + |x| = |z|\), you get \((x - z)^2 + 2*|x - z|*|x| + x^2 = z^2\).

Case 1: If y > 0, then z < x < 0 => x - z -x = -z => 0 = 0
Case 2: If y < 0, then z > x > 0 => -x + z + x = z => 0 = 0

Poor Quality Q: Even without the statements, the equation is sufficient.

1) z < x => Case 1 => Sufficient
2) y > 0 => Case 1 => Sufficient

ANSWER: D

If zy < xy < 0, is |x - z | + |x| = |z|?

(1) z < x
z - x < 0
zy < xy
(z-x)y < 0
y>0; z<x<0
z---------x---------0------------y
|x-z| = |z| - |x|
SUFFICIENT

(2) y > 0
x<0; z<0
(z-x)y < 0
z-x<0
z<x
z----------x------------0-------------y
|x-z| = |z| - |x|
SUFFICIENT

IMO D

this problem is already "sufficient", even without EITHER of the two statements!
yes, you read that correctly.
proof:

* "zy < xy < 0" means that z and y have opposite signs, and x and y have opposite signs. therefore, x and z have the same sign.
furthermore, z must be farther away from zero than x (because the magnitude of zy is greater than the magnitude of xy).
therefore, there are only 2 possibilities (shown on number line):

y-------0-------x--------z
or

z-------x-------0-------y

now let's turn to the problem statement.
|x - z| is the distance between x and z.
|x| is the distance between 0 and x.
|z| is the distance between 0 and z.

using these interpretations, it's plain that |x - z| + |x| = |z| is ALREADY true for both of these statements.
neither of statements (1) and (2) is necessary.

Is my reasoning correct? Bunuel generis