If y ≠ z, is ((x − y) + (z − y)(/(z − y) = 2 ?

Question

If y ≠ z, is  \frac{(x-y)+(z-y)}{z-y}=2 ?

(1) x - y = 2
(2) x = z

yashikaaggarwal · Answer

To prove: (X-Y)+(Z-Y)/(Z-Y) = 2?

Statement 1: X-Y= 2
Putting value in statement.
2+Z-Y/(Z-Y) = 2
2+Z-Y = 2Z-2Y
2+Y = Z
No value given for Z
(Insufficient)

Statement 2: X=Z
Putting value in statement
=> 2X-2Y/X-Y = 2
=> 2(X-Y)/(X-Y) = 2
=> 2 = 2
(Sufficient)

IMO B

Posted from my mobile device

CrackverbalGMAT · Answer

Restructuring the question, we are trying to find if [(x - y) + (z - y) / z - y = 2
or (x - y) / (z-y) + (z - y) / (z - y) = 2
or (x - y) / (z-y) + 1 = 2
or if (x - y) / (z-y) = 1

From Statement (1) we are given that x - y = 2. 
Therefore 2 / (z - y) = 1 or z - y = 2 Therefore this statement is insufficient

From Statement (2) we are given that x = z
Therefore (z - y) / (z - y) = 1. Therefore Statement B is sufficient

Arun Kumar

chetan2u · Answer

The first step in such questions is to simplify the equation.

\frac{(x-y)+(z-y)}{z-y}=2 
 (x-y)+(z-y)=2(z-y)........x+z-2y=2z-2y.......x=z

So the question finally becomes -- Is x=z?

(1) x - y = 2
Nothing about z

(2) x = z
exact information that we are looking for.

B

mageauler · Answer

Hello!
In statment 1 
2+Z-Y/(Z-Y) = 2
2+Z-Y = 2Z-2Y
2+Y = Z
z-y=2, SO ((x-y)+(z-y))/(z-y) = (2+2)/2 = 2 Suf! No?

mageauler · Answer

Hello!
From statment 1:"x-y=2"
(2+(z-y))/(z-y)
2+z-y=2z-2y
"z-y=2".... SO (2+2)/2=2.... SUF! No?

If y ≠ z, is ((x − y) + (z − y)(/(z − y) = 2 ?

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