On the number line, the distance between x and y is greater than the

Best technique is to draw and visualize their positions in the numberline...

1. xyz < 0
Try x and y and z as all negative...
<---(z)---x--(z)---y-----0------------>

Just this scenario is already giving us two possibilities...
INSUFFICIENT

2. xy < 0
This means x and y has opposite signs...

<----(z)---x--(z)--0-------y------>

Just this scenario is already giving us two possibilities...
INSUFFICIENT

Now, let us combine...
If x and y has opposite signs then for xyz to be negative z must be positive...

scenario 1: <---------x-----0-(z)-------y--------------> YES z is in between
scenario 2: <---------y-----0-----(z)-------x------(z)-------> NO z is not in between

STILL INSUFFICIENT

Answer: E

Bunuel, I have a question. Among the 4 cases why are C & D a "no case".

Because in cases C and D z does not lie between x and y. Remember that the questions asks whether z lies between x and y on a number line.

Bunuel, couldn't it also be possible that the variables are arranged so x is between z and y? For example z->x->y, instead of only x->y->z? It's kind of an incidental point because I got it right, but I want to make sure I am figuring it up the proper way.

Also, when the statements xyz or xy, is that to say how the numbers are located with reference to zero, or is it saying to multiply the numbers?

We know that the distance between x and y is greater than the distance between x and z. This can happen in 4 ways, shown in my post. You can see there that x CAN be between z and y in cases C and D.

As for xyz it means x*y*z.

Using (I) and (II) alone, it is easy to figure that each statement alone is not sufficient.

Combining (I) and (II), we have xyz < 0 and xy < 0, this implies that z is positive
(if z is negative then xyz is > 0 [since xy < 0 as per (II)] and xyz > 0 violates (I). Hence z cannot be negative).
When combining (I) and (II) we can take two cases
x is -ve or y is -ve [both cannot have the same sign as this violates (II)].

We will have the below 3 cases for x is -ve and y is -ve. No single solution exists and hence both together not sufficient. Answer is E.

Thanks

Scenario A ) where XYZ < 0.

2 positives 1 negative
3 negatives.

For three negatives on the number the order can be z, x, y, 0...or x, z, y, 0.

So A is insufficient.

Scenario B) hwere XY < 0.

One negative one positive. So the order can go. x 0 z y...or z x 0 y. So B is insufficient.

Scenario A+B.

If X and Y are both of oppositie signs (XY<0)..then Z has to be postive in order for XYZ < 0.

So One Negative and two positives.

Order can be.....x 0 z y...or y 0 x z.

Thus E.

Here is another way to look at the problem.

According to the question, the distance on x and y is greater than x and z, since we have x as the reference point for calculating the distance.

We can only have two cases.

Case 1 - When x is in the middle of y and z

Case 2 - When x is at the extreme end i.e. on one side.

Statement 1 and Statement 2 doesn't provide us the exact values of x,y,z, therefore, Answer is E

We are given that on the number line, the distance between x and y is greater than the distance between x and z. We need to determine whether z lies between x and y on the number line.

Statement One Alone:

xyz < 0

Using the information in statement one, we have two possible cases:

Case 1: Exactly one variable (either x, y, or z) is negative

Case 2: All three variables are negative.

Even with this information, we cannot determine whether z lies between x and y.

For example, for Case 1, if x = -1, y = 2, and z = 1, then z falls between x and y. However, for Case 2, if x = 1, y = 4, and z = -1, then z does not fall between x and y. Statement one alone is not sufficient. We can eliminate answer choices A and D.

Statement Two Alone:

xy < 0

Using the information from statement two, we know that exactly one of the values x or y is negative and the other is positive. However, without knowing anything about z, we cannot determine whether z falls between x and y. Statement two alone is not sufficient. We can eliminate answer choice B.

Statements One and Two Together:

From the information in statements one and two, we know that z must be positive and exactly one of the values x or y is negative. However, we still we cannot determine whether z falls between x and y or outside x and y.

For example, if x = -1, y = 2, and z = 1, then z falls between x and y. However, if x = 1, y = -2, and z = 2, then z does not fall between x and y. The two statements together are still not sufficient.

Answer: E

Hi All,

We're told that, on a number line, the distance between X and Y is GREATER than the distance between X and Z. We're asked if Z is BETWEEN X and Y on the number line. This is a YES/NO question. This question can be solved in a number of different ways. You can use a combination of TESTing VALUES and Number Properties to get to the solution:

Fact 1: (X)(Y)(Z) < 0

This tells us that the three variables are either all negative OR 1 negative and 2 positives.

IF...
X = 1
Y = -2
Z = 2
Then the answer to the question is NO.

IF...
X = 2
Y = -2
Z = 1
Then the answer to the question is YES.
Fact 1 is INSUFFICIENT

Fact 2: (X)(Y) < 0

This tells us that one variable is positive and the other is negative. Unfortunately it tells us NOTHING about Z. As it stands though, the TESTs that I did in Fact 1 fit Fact 2 as well....

IF...
X = 1
Y = -2
Z = 2
Then the answer to the question is NO.

IF...
X = 2
Y = -2
Z = 1
Then the answer to the question is YES.
Fact 2 is INSUFFICIENT

Combined, we already have two TESTs that fit BOTH Facts and produce different answers (a NO and a YES), so there's no more work needed.
Combined, INSUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

From the information, we know that |x-y|>|x-z|, now we need to determine whether z is between x and y.

(1) xyz < 0
This tells us that all 3 of the values are negative (which could mean that z is in between x and y or not) or z is negative and x and y are positive
Insufficient

(2) xy < 0
This tells us that x and y have opposite signs, but tells nothing about z
Insufficient

1+2
Because we know that either x or y has to be negative (from statement 2), that means Z HAS to be negative. But Z could be greater than X (which would give a yes), or Z could be is less than X (which would give a no)

Both Statements are insufficient.
Answer is E

(1) xyz < 0 , atleast one or all of the 3 is negative.Too broad to define. it can be like (x = -4, z = -1, y = -10) (x = 2, y = 6, z =-1)(z is the rightmost and leftmost respectively in the number line). or ( x= 6, y = -1, z = 1)(here z is in the middle of x and y). Not sufficient.

(2) xy < 0. either x or y has to be negative. No information about z. Not sufficient.

Together, lets take x = 5, y = -2, z = 6 ( z is the rightmost), or x = -8, y = 2 and z = -7 (z in the middle). so Not sufficient.

E is the answer.

On the number line the distance bet x & y is greater than the distance between x & z.
^^ given this statement, the number line can appear in one of exactly four ways:

case 1:--------------------x------z-------y---------
case 2: -------------z------x--------------y---------
case 3: -----y--------------x------z-----------------
case 4: -----y------z-------x------------------------

cases 1 and 4 are "yes" to the question; cases 2 and 3 are "no". since the GOAL is to TRY FOR "INSUFFICIENT" (as in just about any non-algebraic DS problem), you should try to get one instance of case 1 or 4, and one instance of case 2 or 3.

(1) xyz<0
^^ you can make any of the four cases with 3 negative numbers, so this statement is insufficient (you don't even have to consider the cases in which some of them are positive).
e.g.
x = -3, z = -2, y = -1 --> case 1 ("yes" to the question)
z = -4, x = -3, y = -1 --> case 2 ("no" to the question)

(2) xy<0
^^ again, easy to get any of the three cases -- x and y just have to be one positive and one negative, and z can be, well, anything at all.
e.g.
x = -2, z = 1, y = 2 --> case 1 ("yes" to the question)
z = -3, x = -2, y = 2 --> case 2 ("no" to the question)

together:
if you have both statements, then you know:
x and y are one positive, one negative;
z is positive.
you can still get both answers:
x = -2, z = 1, y = 2 --> case 1 ("yes" to the question)
y = -2, x = 2, z = 3 --> case 2 ("no" to the question)
still insufficient.

answer = (e)

(1)Option (a) ----z---0-x-------------y--- = xyz<0; Z IS OUTSIDE X AND Y
Option (b) ------x--0--z-----------y---- =xyz<0; Z IS IN BETWEEN X AND Y
Option (c) ------x---z--------y----0---- =xyz<0; Z IS IN BETWEEN X AND Y
Option (d)-------z---x--------y----0---- =xyz<0; Z IS OUTSIDE X AND Y

INSUFFICIENT

(2) Option (A) ------x--------0-------y-= xy<0; Z could be anywhere on the number line. INSUFFICIENT

Using both the information.
No additional information is available to give a confirm point for Z.

Ans. E.

Bunuel

Can I say that because the question talks about only distance/magnitude and not signs and the options talk about only signs and not distance the answer would be E?

Bunuel

Source: Gmat-Prep