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xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z|

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xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z| [#permalink]

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[GMAT math practice question]

xz < 0?

1) |x+y| = |x| + |y|
2) |y+z| = |y| + |z|
[Reveal] Spoiler: OA

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Last edited by MathRevolution on 06 Nov 2017, 22:07, edited 3 times in total.

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Re: xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z| [#permalink]

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New post 31 Oct 2017, 03:22
chetan2u wrote:
MathRevolution wrote:
[GMAT math practice question]

xz < 0?

1) |x+y| = |x| + |y|
2) |y+z| = |y| + |z|



Hi...

The OA is wrong.
Pl rectify


Can you please help solve this

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Re: xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z| [#permalink]

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New post 31 Oct 2017, 05:35
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Expert's post
MathRevolution wrote:
[GMAT math practice question]

xz < 0?

1) |x+y| = |x| + |y|
2) |y+z| = |y| + |z|


HI rahulkashyap,
OA given is A, maximum students would miss out on a crucial point and ans C but the answer should be E.

xz < 0 MEANS " Is x and z of OPPOSITE sign?"

1) |x+y| = |x| + |y|
This tells us that x and y are of SAME sign..
But what about z?
Insuff

2) |y+z| = |y| + |z|
This tells us that z and y are of SAME sign..
But what about x?
Insuff

combined..
x,y and Z are of SAME sign, so xz>0..
so should be sufficient..


BUT what if y is 0
1) |x+y| = |x| + |y|..... y = 0...... x can be POSITIVE or NEGATIVE
2) |y+z| = |y| + |z|..... y = 0...... z can be POSITIVE or NEGATIVE
so x and z can be ANY sign
Insuff


E
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Re: xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z| [#permalink]

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New post 31 Oct 2017, 06:40
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chetan2u wrote:
MathRevolution wrote:
[GMAT math practice question]

xz < 0?

1) |x+y| = |x| + |y|
2) |y+z| = |y| + |z|


HI rahulkashyap,
OA given is A, max would miss out on a ccrucial point and ans C but the answer should be E.

xz < 0 MEANS " Is x and z of OPPOSITE sign?"

1) |x+y| = |x| + |y|
This tells us that x and y are of SAME sign..
But what about z?
Insuff

2) |y+z| = |y| + |z|
This tells us that z and y are of SAME sign..
But what about x?
Insuff

combined..
x,y and Z are of SAME sign, so xz>0..
so should be sufficient..


BUT what if y is 0
1) |x+y| = |x| + |y|..... y = 0...... x can be POSITIVE or NEGATIVE
2) |y+z| = |y| + |z|..... y = 0...... z can be POSITIVE or NEGATIVE
so x and z can be ANY sign
Insuff


E

Amazing skill displayed by pointing out 'y=0' case.
+1.
Thanks chetan2u
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Re: xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z| [#permalink]

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New post 31 Oct 2017, 11:44
MathRevolution wrote:
[GMAT math practice question]

xz < 0?

1) |x+y| = |x| + |y|
2) |y+z| = |y| + |z|


I got this wrong. I need to remember these problems can always be zero.
_________________

I'd love to hear any feedback or ways to improve my problem solving. I make a lot of silly mistakes. If you've had luck improving on stupid mistakes, I'd love to hear how you did it.

Also, I appreciate any kudos.

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Re: xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z| [#permalink]

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New post 01 Nov 2017, 05:13
1) |x+y| = |x| + |y|
This tells us that x and y are of SAME sign..

Can you please explain, How x and y have same signs

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Re: xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z| [#permalink]

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New post 02 Nov 2017, 00:04
=>
Forget conventional ways of solving math questions. In DS, VA (Variable Approach) method is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures a solution.

Since we have 3 variables and 0 equation, E is most likely to be the answer.

Conditions 1) & 2)

From the condition 1) and 2),
|x+y| = |x| + |y| is equivalent to xy ≥ 0 and |y+z| = |y| + |z| is equivalent to yz ≥ 0. Then we have xy^2z ≥ 0 or xz ≥ 0.

x = 1, y = 0, z = 1 implies xz = 1 > 0.
x = -1, y = 0, z = 1 implies xz = -1 < 0.

Both conditions together are not sufficient.

The answer is E.

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80 % chance that E is the answer, while C has 15% chance and A, B or D has 5% chance. Thus, E is most likely to be the answer using 1) and 2) together according to DS definition. Obviously, there may be cases where the answer is A, B, C or D.
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Last edited by MathRevolution on 06 Nov 2017, 22:07, edited 2 times in total.

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Re: xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z| [#permalink]

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New post 02 Nov 2017, 05:42
chetan2u wrote:
MathRevolution wrote:
[GMAT math practice question]

xz < 0?

1) |x+y| = |x| + |y|
2) |y+z| = |y| + |z|


HI rahulkashyap,
OA given is A, maximum students would miss out on a crucial point and ans C but the answer should be E.

xz < 0 MEANS " Is x and z of OPPOSITE sign?"

1) |x+y| = |x| + |y|
This tells us that x and y are of SAME sign..
But what about z?
Insuff

2) |y+z| = |y| + |z|
This tells us that z and y are of SAME sign..
But what about x?
Insuff

combined..
x,y and Z are of SAME sign, so xz>0..
so should be sufficient..


BUT what if y is 0
1) |x+y| = |x| + |y|..... y = 0...... x can be POSITIVE or NEGATIVE
2) |y+z| = |y| + |z|..... y = 0...... z can be POSITIVE or NEGATIVE
so x and z can be ANY sign
Insuff


E


For the answer to be E, even if x or z=0, that would work, correct?

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Re: xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z| [#permalink]

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New post 03 Nov 2017, 19:01
MathRevolution wrote:
[GMAT math practice question]

xz < 0?

1) |x+y| = |x| + |y|
2) |y+z| = |y| + |z|


Statmnt 1: This is possible only when x,y = +ve or x,y = -ve numer
But we dont know about Z .. so insuff

Statmnt 2: This is possible only when y,z = +ve or y,z = -ve
But we dont know X.. so insuff..

Combining above 2 statmnts.. Insufff
For Ex - when x,y is +ve... y,z can be +ve or - ve..

Please give +1 Kudos if you liked my Solution.. :-)

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Re: xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z| [#permalink]

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New post 03 Nov 2017, 19:24
nishantmba wrote:
1) |x+y| = |x| + |y|
This tells us that x and y are of SAME sign..

Can you please explain, How x and y have same signs


Hi nishantmba

RHS is the summation of two positive numbers i.e |x|+|y|, so LHS must be equal to this.
this is only possible when both are of equal sign i.e {+,+} or {-, -} or x=y=0. If either of the two numbers have different sign i.e{+,-}, then |x+y| will yield a less value and hence will not be equal to RHS

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Re: xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z| [#permalink]

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New post 04 Nov 2017, 10:09
Plunkster82 wrote:
MathRevolution wrote:
[GMAT math practice question]

xz < 0?

1) |x+y| = |x| + |y|
2) |y+z| = |y| + |z|


I got this wrong. I need to remember these problems can always be zero.


Hey,

Would you please share your knowledge about "these" problems? What pattern in questions are you referring to?
I have got the idea but want to get extra sure from you.
Thank you
_________________

------------------------------
"Trust the timing of your life"
Hit Kudus if this has helped you get closer to your goal, and also to assist others save time. Tq :)

Kudos [?]: 91 [0], given: 465

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Re: xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z| [#permalink]

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New post 10 Nov 2017, 08:05
TaN1213 wrote:
Plunkster82 wrote:
MathRevolution wrote:
[GMAT math practice question]

xz < 0?

1) |x+y| = |x| + |y|
2) |y+z| = |y| + |z|


I got this wrong. I need to remember these problems can always be zero.


Hey,

Would you please share your knowledge about "these" problems? What pattern in questions are you referring to?
I have got the idea but want to get extra sure from you.
Thank you



The point of this question is the following property.

\(|x+y| = |x| + |y|\) is equivalent to \(xy ≥ 0\).
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The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
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“Receive 5 Math Questions & Solutions Daily”
Unlimited Access to over 120 free video lessons - try it yourself
See our Youtube demo

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Re: xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z| [#permalink]

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New post 10 Nov 2017, 08:08
niks18 wrote:
nishantmba wrote:
1) |x+y| = |x| + |y|
This tells us that x and y are of SAME sign..

Can you please explain, How x and y have same signs


Hi nishantmba

RHS is the summation of two positive numbers i.e |x|+|y|, so LHS must be equal to this.
this is only possible when both are of equal sign i.e {+,+} or {-, -} or x=y=0. If either of the two numbers have different sign i.e{+,-}, then |x+y| will yield a less value and hence will not be equal to RHS



\(|x+y| = |x| + |y|\)
\(⇔ |x+y|^2 = (|x| + |y|)^2\)
\(⇔ (x+y)^2 = |x|^2 + 2|x||y| + |y|^2\)
\(⇔ x^2 + 2xy + y^2 = x^2 + 2|xy| + y^2\)
\(⇔ 2xy = 2|xy|\)
\(⇔ xy = |xy|\)
\(⇔ xy ≥ 0\)
_________________

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The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
Find a 10% off coupon code for GMAT Club members.
“Receive 5 Math Questions & Solutions Daily”
Unlimited Access to over 120 free video lessons - try it yourself
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Re: xz < 0? 1) |x+y| = |x| + |y| 2) |y+z| = |y| + |z|   [#permalink] 10 Nov 2017, 08:08
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