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# Whether x>y>z?

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Manager
Joined: 05 Dec 2009
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12 Apr 2010, 14:21
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Question Stats:

52% (01:54) correct 48% (01:36) wrong based on 396 sessions

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Whether x>y>z?

(1) x-y = |x-z|+|z-y|
(2) x > y
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Joined: 02 Sep 2009
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13 Apr 2010, 02:41
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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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Senior Manager
Joined: 24 Jul 2009
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12 Apr 2010, 16:35
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.
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13 Apr 2010, 04:35
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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13 Apr 2010, 05:27
2
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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24 Sep 2010, 00:18
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

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24 Sep 2010, 00:36
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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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24 Sep 2010, 11:59
Bunnel, thanks as always for the explanation! was considering (x-z) as a negative number..It's clear now..
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24 Sep 2010, 15:44
great explanation.

i think the algebraic method is easier to understand in this one.
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01 Jun 2011, 06:27
1
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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06 Jun 2011, 03:07
good question indeed.

assuming the stem to be correct and then solving A is a cool approach.
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06 Jun 2011, 18:55
I agree with slightly different approach for solving. You rock Bunuel!
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10 Jun 2011, 12:56
2
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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09 Oct 2013, 00:06
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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14 Apr 2014, 09:59
what about the scenario when X=Y=Z
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14 Apr 2014, 10:04
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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