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

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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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New post 14 Jun 2016, 04:38
Bunuel wrote:
This is not a good question, (well at least strange enough) as neither of statement is needed to answer the question, stem is enough to do so. This is the only question from the official source where the statements aren't needed to answer the question. I doubt that such a question will occur on real test but if it ever happens then the answer would be D.

If \(zy<xy<0\) is \(|x-z|+|x| = |z|\)

Look at the inequality \(zy<xy<0\):

We can have two cases:

A. If \(y<0\) --> when reducing we should flip signs and we'll get: \(z>x>0\).
In this case: as \(z>x\) --> \(|x-z|=-x+z\); as \(x>0\) and \(z>0\) --> \(|x|=x\) and \(|z|=z\).

Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(-x+z+x=z\) --> \(0=0\), which is true.

And:

B. If \(y>0\) --> when reducing we'll get: \(z<x<0\).
In this case: as \(z<x\) --> \(|x-z|=x-z\); as \(x<0\) and \(z<0\) --> \(|x|=-x\) and \(|z|=-z\).

Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(x-z-x=-z\) --> \(0=0\), which is true.


So knowing that \(zy<xy<0\) is true, we can conclude that \(|x-z|+|x| = |z|\) will also be true. Answer should be D even not considering the statements themselves.

As for the statements:

Statement (1) says that \(z<x\), hence we have case B.

Statement (2) says that \(y>0\), again we have case B.

Answer: D.

Hope it helps.


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?

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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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New post 14 Jun 2016, 05:59
nishatfarhat87 wrote:
Bunuel wrote:
This is not a good question, (well at least strange enough) as neither of statement is needed to answer the question, stem is enough to do so. This is the only question from the official source where the statements aren't needed to answer the question. I doubt that such a question will occur on real test but if it ever happens then the answer would be D.

If \(zy<xy<0\) is \(|x-z|+|x| = |z|\)

Look at the inequality \(zy<xy<0\):

We can have two cases:

A. If \(y<0\) --> when reducing we should flip signs and we'll get: \(z>x>0\).
In this case: as \(z>x\) --> \(|x-z|=-x+z\); as \(x>0\) and \(z>0\) --> \(|x|=x\) and \(|z|=z\).

Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(-x+z+x=z\) --> \(0=0\), which is true.

And:

B. If \(y>0\) --> when reducing we'll get: \(z<x<0\).
In this case: as \(z<x\) --> \(|x-z|=x-z\); as \(x<0\) and \(z<0\) --> \(|x|=-x\) and \(|z|=-z\).

Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(x-z-x=-z\) --> \(0=0\), which is true.


So knowing that \(zy<xy<0\) is true, we can conclude that \(|x-z|+|x| = |z|\) will also be true. Answer should be D even not considering the statements themselves.

As for the statements:

Statement (1) says that \(z<x\), hence we have case B.

Statement (2) says that \(y>0\), again we have case B.

Answer: D.

Hope it helps.


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.
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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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New post 13 Aug 2016, 12:34
I am not really sure what the question is, because the question stem answers the question and Statements 1 and 2 are useless to answer the question.

The question is: Is |X-Z| + |X| = |Z|?

In other words: Is |X-Z| = |Z| - |X|?

Since |X-Z| must be positive (not zero because ZY<XY<0), the question becomes:Is |Z|>|X|?

In this sense, the actual question is:
Is: Z<X<0 ?
or is 0<X<Z ?

However, the statements are useless, because from the question stem we already know that since ZY<XY<0
If Y is +ve, Z<X<0
If Y is -ve, 0<X<Z
In other words |Z|>|X|

The statements merely confirm what we already know..

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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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New post 09 Oct 2016, 21:16
I like to maximize my usage of number plugging. Here is my method:
ZY<XY<0 => ZY - XY < 0 => Y*(Z-X) < 0
(i) Y<0 => Z-X>0 => Z>X. Also, ZY<0 => Z>0 & X>0 => Z>X>0
&
(ii) Y>0 => Z-X<0 => Z<X. Also, XY<0 => X<0 & Z<0 => Z<X<0

Also, |X-Z|+|X|=|Z| => |X-Z| = |Z| - |X|

STATEMENT1: Z<X => Z<X<0 => -2<-1<0 => |-1+2| = 1 & |-2| - |-1| = 2-1 = 1 => 1 = 1. Sufficient

STATEMENT 2: Y<0 => Z>X>0 => 2>1>0 => |1-2| = |-1| = 1 & |2| - |1| = 2-1 = 1 => 1 = 1. Sufficient.
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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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New post 31 Oct 2016, 12:38
VeritasPrepKarishma wrote:
rgtiwari wrote:
If zy < xy < 0, is | x - z | + |x| = |z|?

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


You can solve such questions easily by re-stating '< 0' as 'negative' and '> 0' as 'positive'.

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.

When will zy and xy both be negative? In 2 cases:
Case 1: When y is positive and z and x are both negative.
Case 2: When y is negative and z and x are both positive.

Question:
Is | x - z | + |x| = |z| ?
Is | x - z | = |z| - |x| ?
Is | z - x | = |z| - |x| ? (Since | x - z | = | z - x |)
We re-write the question only for better understanding.

Now think, when will | z - x | = |z| - |x| ? It happens when x and z have the same sign. In case both have the same sign, they get subtracted on both sides so you get the same answer. In case they have opposite signs, they get added on LHS and subtracted on RHS and hence the equality doesn't hold.

So if we can figure whether both x and z have the same sign, we can answer the question.

As we saw above, in both case 1 and case 2, x and z must have the same sign. This implies that the equality must hold and you don't actually need the statements to answer the question. You can answer it without the statements (this shouldn't happen in actual GMAT).
Hence answer must be (D).


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

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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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New post 01 Nov 2016, 01:27
yezz wrote:
VeritasPrepKarishma wrote:
rgtiwari wrote:
If zy < xy < 0, is | x - z | + |x| = |z|?

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


You can solve such questions easily by re-stating '< 0' as 'negative' and '> 0' as 'positive'.

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.

When will zy and xy both be negative? In 2 cases:
Case 1: When y is positive and z and x are both negative.
Case 2: When y is negative and z and x are both positive.

Question:
Is | x - z | + |x| = |z| ?
Is | x - z | = |z| - |x| ?
Is | z - x | = |z| - |x| ? (Since | x - z | = | z - x |)
We re-write the question only for better understanding.

Now think, when will | z - x | = |z| - |x| ? It happens when x and z have the same sign. In case both have the same sign, they get subtracted on both sides so you get the same answer. In case they have opposite signs, they get added on LHS and subtracted on RHS and hence the equality doesn't hold.

So if we can figure whether both x and z have the same sign, we can answer the question.

As we saw above, in both case 1 and case 2, x and z must have the same sign. This implies that the equality must hold and you don't actually need the statements to answer the question. You can answer it without the statements (this shouldn't happen in actual GMAT).
Hence answer must be (D).


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."
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If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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New post 25 Feb 2017, 12:11
Bunuel wrote:
This is not a good question, (well at least strange enough) as neither of statement is needed to answer the question, stem is enough to do so. This is the only question from the official source where the statements aren't needed to answer the question. I doubt that such a question will occur on real test but if it ever happens then the answer would be D.

If \(zy<xy<0\) is \(|x-z|+|x| = |z|\)

Look at the inequality \(zy<xy<0\):

We can have two cases:

A. If \(y<0\) --> when reducing we should flip signs and we'll get: \(z>x>0\).
In this case: as \(z>x\) --> \(|x-z|=-x+z\); as \(x>0\) and \(z>0\) --> \(|x|=x\) and \(|z|=z\).

Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(-x+z+x=z\) --> \(0=0\), which is true.

And:

B. If \(y>0\) --> when reducing we'll get: \(z<x<0\).
In this case: as \(z<x\) --> \(|x-z|=x-z\); as \(x<0\) and \(z<0\) --> \(|x|=-x\) and \(|z|=-z\).

Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(x-z-x=-z\) --> \(0=0\), which is true.


So knowing that \(zy<xy<0\) is true, we can conclude that \(|x-z|+|x| = |z|\) will also be true. Answer should be D even not considering the statements themselves.

As for the statements:

Statement (1) says that \(z<x\), hence we have case B.

Statement (2) says that \(y>0\), again we have case B.

Answer: D.

Hope it helps.



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?

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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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OreoShake wrote:
Bunuel wrote:
This is not a good question, (well at least strange enough) as neither of statement is needed to answer the question, stem is enough to do so. This is the only question from the official source where the statements aren't needed to answer the question. I doubt that such a question will occur on real test but if it ever happens then the answer would be D.

If \(zy<xy<0\) is \(|x-z|+|x| = |z|\)

Look at the inequality \(zy<xy<0\):

We can have two cases:

A. If \(y<0\) --> when reducing we should flip signs and we'll get: \(z>x>0\).
In this case: as \(z>x\) --> \(|x-z|=-x+z\); as \(x>0\) and \(z>0\) --> \(|x|=x\) and \(|z|=z\).

Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(-x+z+x=z\) --> \(0=0\), which is true.

And:

B. If \(y>0\) --> when reducing we'll get: \(z<x<0\).
In this case: as \(z<x\) --> \(|x-z|=x-z\); as \(x<0\) and \(z<0\) --> \(|x|=-x\) and \(|z|=-z\).

Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(x-z-x=-z\) --> \(0=0\), which is true.


So knowing that \(zy<xy<0\) is true, we can conclude that \(|x-z|+|x| = |z|\) will also be true. Answer should be D even not considering the statements themselves.

As for the statements:

Statement (1) says that \(z<x\), hence we have case B.

Statement (2) says that \(y>0\), again we have case B.

Answer: D.

Hope it helps.



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 \(|z|=z\).
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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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New post 13 Dec 2017, 19:58
Bunuel wrote:
This is not a good question, (well at least strange enough) as neither of statement is needed to answer the question, stem is enough to do so. This is the only question from the official source where the statements aren't needed to answer the question. I doubt that such a question will occur on real test but if it ever happens then the answer would be D.

If \(zy<xy<0\) is \(|x-z|+|x| = |z|\)

Look at the inequality \(zy<xy<0\):

We can have two cases:

A. If \(y<0\) --> when reducing we should flip signs and we'll get: \(z>x>0\).
In this case: as \(z>x\) --> \(|x-z|=-x+z\); as \(x>0\) and \(z>0\) --> \(|x|=x\) and \(|z|=z\).

Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(-x+z+x=z\) --> \(0=0\), which is true.

And:

B. If \(y>0\) --> when reducing we'll get: \(z<x<0\).
In this case: as \(z<x\) --> \(|x-z|=x-z\); as \(x<0\) and \(z<0\) --> \(|x|=-x\) and \(|z|=-z\).

Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(x-z-x=-z\) --> \(0=0\), which is true.


So knowing that \(zy<xy<0\) is true, we can conclude that \(|x-z|+|x| = |z|\) will also be true. Answer should be D even not considering the statements themselves.

As for the statements:

Statement (1) says that \(z<x\), hence we have case B.

Statement (2) says that \(y>0\), again we have case B.

Answer: D.

Hope it helps.




Hi Buneul,

Does the below explanation make sense?

since |x-z| + |x|=|z|

We can break this down to

1) (z)=(x-z)+(x)
2) (z)=-(x-z)-(x)

Hence,
1) z=2x-z --> z=x
2) x=0

In both cases, |x-z| + |x|=|z| holds true and hence sufficient.

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Re: If zy < xy < 0, is | x - z | + |x| = |z|? [#permalink]

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New post 23 Dec 2017, 10:49
rgtiwari wrote:
If zy < xy < 0, is | x - z | + |x| = |z|?

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




To think alternatively this is a very easy question hardly a minute,
Given IF zy<xy<0
this means that both zy and xy have opposite signs
There can be two cases Either y is positive then both z and x are negative----- Here z<x----Z-X<0
Cae 2 y is negative . Here both x and y are positive and for inequality to be true -------now x<z----Z-X>0

Rewriting
IS | x - z | + |x| = |z|,,,=
IS(((| z-x | = |z| - |x|))) Which basically means that we have to find whether case 2 hold or CASE 1

St 1 Z<X ------
Case 1 holds here and therefore | z-x | is equal to |z| - |x| ((((CASE 1 Z-X<0))) you can substitute some numbers to check------ SUFFICIENT(((( Z=-6 , X=-2 ,,,, LHS =4 AND RHS = 6 - 2 =4
ST 2 Y>0 CASE 1 HOLDS AGAIN HENCE SUFFICIENT

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Re: If zy < xy < 0, is | x - z | + |x| = |z|?   [#permalink] 23 Dec 2017, 10:49

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