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

Question

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

pk123 · Accepted Answer

1) 0< x< y 2) x< y Option A: 00, 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

lostboy · Answer

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

raghavshekhar · Answer

A.

Getting first sufficient through number plugging and second insufficient through number plugging.

yenbh · Answer

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

andrenoble · Answer

Solution

We should consider x and y can have different signs (unless we are provided with a different information).

(1) Is sufficient - as described above
(2) x and y could be either both < 0, or both > 0, OR x could be < 0, while y is > 0 - and the last permutation makes (2) insufficient.

Hope is helpful

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

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