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

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

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20 Oct 2018, 10:40
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55% (hard)

Question Stats:

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

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If z > 0, is $$\frac{(x+z)}{(y+z)} > \frac{x}{y}$$?
1) 0< x< y
2) x< y

GMATbuster's Weekly GMAT Quant Quiz #5 Ques No 9

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Joined: 16 Sep 2011
Posts: 97
Re: If z > 0, is [m][fraction](x+z)/(y+z)[/fraction] > [fraction]x/y[/frac  [#permalink]

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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
Manager
Joined: 02 Apr 2018
Posts: 50
Re: If z > 0, is [m][fraction](x+z)/(y+z)[/fraction] > [fraction]x/y[/frac  [#permalink]

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20 Oct 2018, 10:49

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
Intern
Joined: 08 Feb 2018
Posts: 20
Re: If z > 0, is [m][fraction](x+z)/(y+z)[/fraction] > [fraction]x/y[/frac  [#permalink]

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20 Oct 2018, 10:58
A.

Getting first sufficient through number plugging and second insufficient through number plugging.
Intern
Joined: 17 Dec 2016
Posts: 19
Location: Viet Nam
Re: If z > 0, is [m][fraction](x+z)/(y+z)[/fraction] > [fraction]x/y[/frac  [#permalink]

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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

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

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