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If z > 0, is [m][fraction](x+z)/(y+z)[/fraction] > [fraction]x/y[/frac

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Concentration: International Business, General Management
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If z > 0, is [m][fraction](x+z)/(y+z)[/fraction] > [fraction]x/y[/frac  [#permalink]

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New post 20 Oct 2018, 10:40
1
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

59% (01:57) correct 41% (01:37) wrong based on 28 sessions

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Re: If z > 0, is [m][fraction](x+z)/(y+z)[/fraction] > [fraction]x/y[/frac  [#permalink]

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New post 20 Oct 2018, 10:54
1) 0< x< y
2) x< y

Option A: 0<x<y
which says 0<x/y<1

hence lets take x/y=1/2....if you add any value of Z as positive to the positive number , it will always be greater than original

that is (x+z)/(y+z) will always be greater than x/y if z>0,

to illustrate x= 1, y= 2 and z= 0.1 which gives x/y= 1/2= 0.5
but (x+z)/(y+z)= 1.1/2.1> 0.5 as it is euual to 0.55

hence sufficient


Option B
x<y

which means x and y can be negative as well

if x= -2 and y = -1
x/y= 2
if we take z= 0.5, it becomes (x+z)/(y+z)=-1.5/-0.5= 3, which is more than original x/y= 2
but if we take z= 2, it becomes 0/1= 0 which is less

hence not sufficient

A is the answer
General Discussion
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Re: If z > 0, is [m][fraction](x+z)/(y+z)[/fraction] > [fraction]x/y[/frac  [#permalink]

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New post 20 Oct 2018, 10:49
Answer is d:

we do not need to know that x and y are positive you can plug in negative values to test this

let x = -3 and y = -2 and z = 1
therefore
LHS: (-3 + 1) / (-2+1)
=2
RHS: -3/-2
=3/2

Therefore LHS > RHS
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Re: If z > 0, is [m][fraction](x+z)/(y+z)[/fraction] > [fraction]x/y[/frac  [#permalink]

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New post 20 Oct 2018, 10:58
A.

Getting first sufficient through number plugging and second insufficient through number plugging.
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Re: If z > 0, is [m][fraction](x+z)/(y+z)[/fraction] > [fraction]x/y[/frac  [#permalink]

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New post 20 Oct 2018, 11:00
Statement 1: 0< x< y, since z> 0 clearly \(\frac{(x+z)}{(y+z)}\)>\(\frac{x}{y}\)

Take x =1 , y = 2 for example, 1/2<3/4 --> Sufficient

Statement 2: x <y

+ In case 0 <x<y --> Same as statement 1 \(\frac{(x+z)}{(y+z)}\)>\(\frac{x}{y}\)
+ In case x <y<0 --> \(\frac{x}{y}\) > 0, similar to above case, \(\frac{(x+z)}{(y+z)}\)>\(\frac{x}{y}\)
+ In case x <=0<y, take x = -1, y = 2, so: -1/2<(-1+1) + (2+1)
--> can conclude \(\frac{(x+z)}{(y+z)}\)>\(\frac{x}{y}\)
--> Statement 2 is sufficient

Answer: D
GMAT Club Bot
Re: If z > 0, is [m][fraction](x+z)/(y+z)[/fraction] > [fraction]x/y[/frac   [#permalink] 20 Oct 2018, 11:00
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