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Manager  Joined: 05 Dec 2009
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Difficulty:   75% (hard)

Question Stats: 50% (01:48) correct 50% (02:04) wrong based on 350 sessions

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Whether x>y>z?

(1) x-y = |x-z|+|z-y|
(2) x > y
Math Expert V
Joined: 02 Sep 2009
Posts: 59722

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xyztroy wrote:
Whether x>y>z?
1. x-y = |x-z|+|z-y|
2. x > y

Q: is $$x>y>z$$?

(1) $$x-y=|x-z|+|z-y|$$

First of all as RHS is the sum of two non-negative values LHS also must be non-negative. So $$x-y\geq{0}$$.

Now, we are told that the distance between two points $$x$$ and $$y$$, on the number line, equals to the sum of the distances between $$x$$ and $$z$$ AND $$z$$ and $$y$$.

The question is: can the points placed on the number line as follows ---z---y---x---. If you look at the number line you'll see that it's just not possible. Sufficient.

OR algebraic approach:

If $$x>y>z$$ is true, then $$x-y=|x-z|+|z-y|$$, will become $$x-y=x-z-z+y$$ --> $$z=y$$, which contradicts our assumption $$x>y>z$$. So $$x>y>z$$ is not possible. Sufficient.

(2) $$x>y$$ --> no info about $$z$$. Not sufficient.

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Manager  Joined: 24 Jul 2009
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IMHO E

from the given equation.

1. x-y = |x-z|+|z-y|

we can have four cases,

a) x > z > y . The given equations gives no result.
b) x > z and z < y. We get y=z
c) x < z and z > y. We get x=z
d) y > z > x. We get x=y

Though we have 3 (b,c and d)results which implies that x > y > z cannot be true.
But the result from a is not clear. So we cannot confidently state that x > y > z is true. INSUFFICIENT.

2. x > y. INSUFFICIENT.

From 1 and 2.

given x > y.(Statement 2).We can only consider cases a, b and c.(From statement 1).

Again from a, b and c. NOT SUFFICIENT.

OA plz.
Manager  Joined: 24 Jul 2009
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Bunuel wrote:
xyztroy wrote:
Whether x>y>z?
1. x-y = |x-z|+|z-y|
2. x > y

Q: is $$x>y>z$$?

(1) $$x-y=|x-z|+|z-y|$$

First of all as RHS is the sum of two non-negative values LHS also must be non-negative. So $$x-y\geq{0}$$.

Now, we are told that the distance between two points $$x$$ and $$y$$, on the number line, equals to the sum of the distances between $$x$$ and $$z$$ AND $$z$$ and $$y$$.

The question is: can the points placed on the number line as follows ---z---y---x---. If you look at the number line you'll see that it's just not possible. Sufficient.

OR algebraic approach:

If $$x>y>z$$ is true, then $$x-y=|x-z|+|z-y|$$, will become $$x-y=x-z-z+y$$ --> $$z=y$$, which contradicts our assumption $$x>y>z$$. So $$x>y>z$$ is not possible. Sufficient.

(2) $$x>y$$ --> no info about $$z$$. Not sufficient.

hello Brunel..

IMHO E

from the given equation.

1. x-y = |x-z|+|z-y|

we can have four cases,

a) x > z > y . The given equations gives no result.
b) x > z and z < y. We get y=z
c) x < z and z > y. We get x=z
d) y > z > x. We get x=y

Though we have 3 (b,c and d)results which implies that x > y > z cannot be true.
But the result from a is not clear. So we cannot confidently state that x > y > z is true. INSUFFICIENT.

2. x > y. INSUFFICIENT.

From 1 and 2.

given x > y.(Statement 2).We can only consider cases a, b and c.(From statement 1).

Again from a, b and c. NOT SUFFICIENT.
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Joined: 02 Sep 2009
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nverma wrote:

hello Brunel..

IMHO E

from the given equation.

1. x-y = |x-z|+|z-y|

we can have four cases,

a) x > z > y . The given equations gives no result.
b) x > z and z < y. We get y=z
c) x < z and z > y. We get x=z
d) y > z > x. We get x=y

Though we have 3 (b,c and d)results which implies that x > y > z cannot be true.
But the result from a is not clear. So we cannot confidently state that x > y > z is true. INSUFFICIENT.

2. x > y. INSUFFICIENT.

From 1 and 2.

given x > y.(Statement 2).We can only consider cases a, b and c.(From statement 1).

Again from a, b and c. NOT SUFFICIENT.

For (a) you are checking scenario x > z > y. But if we have this case then the answer to the question is x>y>z is still NO.

Let's do this in the way you are solving:

Question: is $$z<y<x$$?

(1) We have $$x-y = |x-z|+|z-y|$$. This statement is true. Now let's check in which cases is this true:

As $$x\geq{y}$$ (see my solution to see why this must be true) then there can be 4 scenarios as you've written:

A. $$x=y$$ --> $$x-y=0=|x-z|+|z-y|$$ --> $$x=y=z$$. Which means that $$x-y = |x-z|+|z-y|$$ is true when $$x=y=z$$.

B. $$z<y<x$$ ---z---y---x--
$$x-y=|x-z|+|z-y|$$ --> $$x-y=x-z-z+y$$ --> $$z=y$$. Which means that $$x-y = |x-z|+|z-y|$$ is also true when $$y={z}<x$$. Two points $$z$$ and $$y$$ must coincide.

C. $$y<z<x$$ ---y---z---x--
$$x-y=|x-z|+|z-y|$$ --> $$x-y=x-z+z-y$$ --> $$0=0$$. Which means that $$x-y = |x-z|+|z-y|$$ is always true for any values of x, y, and z, when $$y<z<x$$. You can see this on diagram: if x, y, and z are placed on number line as above then the distance between two points $$x$$ and $$y$$, on the number line, equals to the sum of the distances between $$x$$ and $$z$$ AND $$z$$ and $$y$$.

D. $$y<x<z$$ ---y---x---z---
$$x-y=|x-z|+|z-y|$$ --> $$x-y=-x+z+z-y$$ --> $$z=x$$. Which means that $$x-y = |x-z|+|z-y|$$ is true when $$y<z=x$$. Two points $$z$$ and $$x$$ must coincide.

So we've got that statement as $$x-y = |x-z|+|z-y|$$ is true, only following 4 scenarios are possible:
A. $$x=y=z$$;
B. $$z=y<x$$;
C. $$y<z<x$$;
D. $$y<z=x$$.

Among the above scenarios there is no case when $$z<y<x$$. Hence the answer to the question "is $$z<y<x$$" is NO.

Hope it's clear.
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Manager  Joined: 16 Jul 2009
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Bunnel, can you please explain why in the algebraic approach for st1 the signs have not been changed for solving x-y = lx-zl + lz-yl , as your solution shows that this equation becomes (after removing the modulus signs) x-y = x-z-z+y --> z=y

Math Expert V
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abhi758 wrote:
Bunnel, can you please explain why in the algebraic approach for st1 the signs have not been changed for solving x-y = lx-zl + lz-yl , as your solution shows that this equation becomes (after removing the modulus signs) x-y = x-z-z+y --> z=y

We know that for $$|x|$$:
When $$x\leq{0}$$, then $$|x|=-x$$;
When $$x\geq{0}$$, then $$|x|=x$$.

So if $$x>y>z$$ is true, then:
As $$x>z$$ ($$x-z>0$$) --> $$|x-z|=x-z$$ AND as $$y>z$$ ($$z-y<0$$) --> $$|z-y|=-(z-y)=-z+y$$.

So $$x-y=|x-z|+|z-y|$$, will become $$x-y=x-z-z+y$$ --> $$z=y$$.

Hope it's clear.
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Manager  Joined: 16 Jul 2009
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Bunnel, thanks as always for the explanation! was considering (x-z) as a negative number..It's clear now..
Manager  Joined: 04 Aug 2010
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great explanation.

i think the algebraic method is easier to understand in this one.
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xyztroy wrote:
Whether x>y>z?
1. x-y = |x-z|+|z-y|
2. x > y

1) x-y = -x+z+z-y
x = z
so x>y>z is not true. Sufficient
2) no information is given about z. Insufficient.
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good question indeed.

assuming the stem to be correct and then solving A is a cool approach.
Manager  Joined: 05 Jul 2010
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I agree with slightly different approach for solving. You rock Bunuel!
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xyztroy wrote:
Whether x>y>z?
1. x-y = |x-z|+|z-y|
2. x > y

Alternative approach:
Stmnt 1: x-y = |x-z|+|z-y|

If the RHS has to be equal to x - y, we have to get rid of 'z'. That will happen only when
EITHER |x-z| = x - z and |z-y| = z - y i.e. x - z and z - y are both positive which implies x > z and z > y
OR |x-z| = -(x - z) and |z-y| = -(z - y) i.e. x - z and z - y are both negative which implies x < z and z < y. (Anyway, in this case, RHS becomes y - x instead of x - y)

In both the cases, the relation between x, y and z is not x>y>z. Hence statement 1 is sufficient to say "No, x>y>z does not hold."
Statement 2 doesn't say anything about z so is obviously not sufficient alone.
Hence A.
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xyztroy wrote:
Whether x>y>z?

(1) x-y = |x-z|+|z-y|
(2) x > y

We need to answer this question in YES or NO.

Statement 2 gives us no information about z, so Insufficient.

Statement 1 :-
x-y >= 0 ; since the RHS term will always be 0 or positive.

Hence, there can be 3 scenarios with 1st statement.

Imagining a number line :-
1) y-------z---------x :- x>z>y and x-y (Distance between x and y) is equal to the distance between x and z & z and y. So False. since y<z
2) z,y ------------- x :- x > z ; x > y ; y = z , Therefore distance between y and z is 0. Still gives us false since y = z and not greater than that.
3) y----------------x,z :- x = z ; x > y ; y < z , Therefore distance between x and z is 0. Still gives us false since y < z.

In all the 2 scenarios we get FALSE. Hence sufficient.

So only A is sufficient.

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what about the scenario when X=Y=Z
Math Expert V
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ongy wrote:
what about the scenario when X=Y=Z

The answer will remain the same. The question asks whether x>y>z. If x=y=z, then the answer to the questions "is x>y>z" is still NO.
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Bunuel wrote:
xyztroy wrote:
Whether x>y>z?
1. x-y = |x-z|+|z-y|
2. x > y

Q: is $$x>y>z$$?

(1) $$x-y=|x-z|+|z-y|$$

First of all as RHS is the sum of two non-negative values LHS also must be non-negative. So $$x-y\geq{0}$$.

Now, we are told that the distance between two points $$x$$ and $$y$$, on the number line, equals to the sum of the distances between $$x$$ and $$z$$ AND $$z$$ and $$y$$.

The question is: can the points placed on the number line as follows ---z---y---x---. If you look at the number line you'll see that it's just not possible. Sufficient.

OR algebraic approach:

If $$x>y>z$$ is true, then $$x-y=|x-z|+|z-y|$$, will become $$x-y=x-z-z+y$$ --> $$z=y$$, which contradicts our assumption $$x>y>z$$. So $$x>y>z$$ is not possible. Sufficient.

(2) $$x>y$$ --> no info about $$z$$. Not sufficient.

Bunuel Can we assume what the question is asking to be true and solve for each statement as done here for statement 1??

Posted from my mobile device Re: Whether x>y>z?   [#permalink] 27 Nov 2019, 01:41
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