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If xy + x = z, is |x + y| > z?

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If xy + x = z, is |x + y| > z?  [#permalink]

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New post 29 Apr 2016, 02:44
1
14
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

65% (02:12) correct 35% (02:03) wrong based on 281 sessions

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Joined: 11 Aug 2013
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Concentration: Finance, General Management
GMAT 1: 620 Q47 V28
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Re: If xy + x = z, is |x + y| > z?  [#permalink]

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New post 29 Apr 2016, 03:32
1
1
Answer: C
Re-writing xy+x=z as x(y+1)=z --(a)
Now looking at the question we can definitely say
|x+y|>z if a) z is -ve as mod of any number is always +ve
OR b) it can be proved that z is positive and > x+y
Taking i) x< 0 our modified equation (a) suggests that sign of z depends on sign of (y+1) so it can be both +ve or -ve. There is no data on the actual numeric value of x ,y or z. so. Not sufficient
Taking ii) y>0 our modified equation (a) suggests that sign of z depends on sign of x so it can be both +ve or -ve. There is no data on the actual numeric value of x ,y or z. so. Not sufficient

Combining i and ii and looking at eq -(a)
x< 0 and y>0 => (y+1)>0. Hence x(y+1) <0 => z <0. From our earlier inference if z<0 then |x+y|>z. Sufficient.
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Re: If xy + x = z, is |x + y| > z?  [#permalink]

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New post 21 Sep 2017, 07:34
neither statement can give us much
but if we combine both here is the interesting thing we can find: whether |x+y|> xy+x

if x is negative and y is positive, then xy will be negative and plus negative will always be negative, and if we sum negative and negative the result will be negative so, xy+x is negative if we combine statements 1) and 2)

|x+y| though will be positive no matter what, so the answer will be a definite yes, so, the answer is C (both 1 &2 are sufficient together)
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Re: If xy + x = z, is |x + y| > z?  [#permalink]

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New post 15 Oct 2017, 00:48
1
By manipulating the condition we get x+y>z? or x+y<-z
Therefore x+y>xy+x => is y(x-1)<0?
x+y<-(xy+x)=> is y(x+1)+2x<0?
We clearly need the signs of x and y
1) No sign of y.Insufficient
2)No sign of X. Insufficient

1)+2)=>Sufficient to prove above statements
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Re: If xy + x = z, is |x + y| > z?  [#permalink]

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New post 15 Oct 2017, 01:33
1
2
Bunuel wrote:
If xy + x = z, is |x+y|>z?

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


Given \(x(y+1)=z\), we need the values of \(x\), \(y\) & \(z\) to determine whether \(|x+y|>z\), where LHS is always positive

Statement 1: implies \(x\) is negative so \(x(y+1)\) i.e. \(z\) can be negative or it can be positive depending upon the value of \((y+1)\). But nothing is mentioned about \(y\). So insufficient


Statement 2: implies \(y\) is positive so \(x(y+1)\) i.e. \(z\) can be negative or it can be positive depending upon the value of \(x\). But nothing is mentioned about \(x\). So insufficient

Combining 1 & 2 we get \(x<0\) & \((y+1)>0\) so \(x(y+1)<0\) i.e. \(z<0\)

but \(|x+y|\) is always positive hence we get a Yes for our question stem. Sufficient

Option C
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If xy + x = z, is |x + y| > z?  [#permalink]

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New post 05 Jun 2018, 03:07
According me E is correct since y is less than zero so y=-1 x= 1 therefore z can be neget ive or +'ve while combining plz correct me if I am wrong
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Re: If xy + x = z, is |x + y| > z?  [#permalink]

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New post 05 Jun 2018, 03:50
Krishaa12 wrote:
According me E is correct since y is less than zero so y=-1 x= 1 therefore z can be neget ive or +'ve while combining plz correct me if I am wrong


Hi Krishaa12

y>0, hence it is positive, rather x<0 i.e negative

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Re: If xy + x = z, is |x + y| > z? &nbs [#permalink] 05 Jun 2018, 03:50
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