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Whether x>y>z?

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Whether x>y>z?  [#permalink]

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New post 12 Apr 2010, 14:21
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Whether x>y>z?

(1) x-y = |x-z|+|z-y|
(2) x > y
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Re: is x>y>z  [#permalink]

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New post 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.

Answer: A.
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Re: is x>y>z  [#permalink]

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New post 12 Apr 2010, 16:35
1
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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Re: is x>y>z  [#permalink]

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New post 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.

Answer: A.



hello Brunel..

Can u please help me with my approach..where i have gone wrong..!!

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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Re: is x>y>z  [#permalink]

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New post 13 Apr 2010, 05:27
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nverma wrote:

hello Brunel..

Can u please help me with my approach..where i have gone wrong..!!

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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Re: is x>y>z  [#permalink]

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New post 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

Look forward to your reply.
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Re: is x>y>z  [#permalink]

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New post 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

Look forward to your reply.


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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Re: is x>y>z  [#permalink]

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New post 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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Re: is x>y>z  [#permalink]

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New post 24 Sep 2010, 15:44
great explanation.

i think the algebraic method is easier to understand in this one.
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Re: is x>y>z  [#permalink]

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New post 01 Jun 2011, 06:27
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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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Re: is x>y>z  [#permalink]

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New post 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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Re: is x>y>z  [#permalink]

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New post 06 Jun 2011, 18:55
I agree with slightly different approach for solving. You rock Bunuel!
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Re: is x>y>z  [#permalink]

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New post 10 Jun 2011, 12:56
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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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Re: Whether x>y>z?  [#permalink]

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New post 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.

Consider Kudos if the post helps anyone. Its a good way to motivate :-D
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Re: Whether x>y>z?  [#permalink]

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New post 14 Apr 2014, 09:59
what about the scenario when X=Y=Z
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Re: Whether x>y>z?  [#permalink]

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New post 14 Apr 2014, 10:04
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Re: Whether x>y>z?  [#permalink]

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New post 27 Nov 2019, 01:41
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.

Answer: A.

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

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Re: Whether x>y>z?   [#permalink] 27 Nov 2019, 01:41
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