If x, y and z are positive integers, is (x - y)(y - z)(x - z) 0?

Question

If x, y and z are positive integers, is  (x - y)(y - z)(x - z) > 0 ?

(1)  x^2 + yz = xy + xz

(2)  xy - y^2 = xz - yz

Viplav Shrivastava · Answer

Answer should be D

1) solve the equation and you get (x-z)(x-y)=0 
Which means either of the terms are zero. Using this inference, (x-y)(y-z)(x-z) = 0. Which is not greater than 0.
Sufficient

2) similarly st. 2 gives (x-y)(y-z)=0
Thus sufficient

Answer-D

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Archit3110 · Answer

#1
 x^2 + yz = xy + xz

x(x-y)=z(x-y)
x=z
so (x - y)(y - z)(x - z)= 0 
sufficient
#2
 xy - y^2 = xz - yz 
solve
y=z
so (x - y)(y - z)(x - z) =0
IMO D

dave13 · Answer

VeritasKarishma  can you solve this (x - y)(y - z)(x - z) > 0

chetan2u · Answer

If x, y and z are positive integers, is  (x - y)(y - z)(x - z) > 0 ?

(1)  x^2 + yz = xy + xz……………x^2+yz-xy-xz=0…………x(x-y)-z(x-y)=(x-z)(x-y)=0 
Thus,  (x - y)(y - z)(x - z) =0 .
Answer is NO
Sufficient

(2)  xy - y^2 = xz - yz……..xy-y^2-xz+yz=0………y(x-y)-z(x-y)=0………(y-z)(x-y)=0 
Thus,  (x - y)(y - z)(x - z) =0 .
Answer is NO
Sufficient

D

Kinshook · Answer

Asked: If x, y and z are positive integers, is  (x - y)(y - z)(x - z) > 0 ?

(1)  x^2 + yz = xy + xz 
x^2 - xz = xy - yz 
x(x-z) = y(x-y)
(x-y)(x-z) = 0
(x - y)(y - z)(x - z) = 0
SUFFICIENT

(2)  xy - y^2 = xz - yz 
y (x-y) = z(x-y)
(x-y)(y-z)= 0
(x - y)(y - z)(x - z) = 0
SUFFICIENT

IMO D

KarishmaB · Answer

You need to do algebraic manipulation here. You start with statements 1 and 2 independently and try to arrive at what is required. x^2 + yz = xy + xz x^2 - xy = xz - yz x(x - y) = z(x - y) x(x - y) - z(x - y) = 0 (x - y)(x - z) = 0 This means x is either equal to y or to z. Then (x - y)(y - z)(x - z) will be equal to 0 because one term will certainly be 0. Solving this: (x - y)(y - z)(x - z) > 0 doesn't make much sense. It's too complicated with 3 variables. (x - y)(y - z)(x - z) > 0 means either all three terms are positive or exactly 1 is positive and the other two are negative. Either (x - y) > 0 and (y - z)>0 and (x - z) > 0 So x > y, y > z and x > z i.e. x > y > z Or (x - y) > 0 and (y - z)<0 and (x - z) < 0 x > y and y < z but x < z too. This is not possible. (x - y) < 0 and (y - z)>0 and (x - z) < 0 x < y and y > z and x < z i.e. x < z < y (x - y) < 0 and (y - z)<0 and (x - z) > 0 x < y and y < z but x > z. This is not possible. So we get two possible relations: x > y > z or x < z < y So now the question is: does either one of these relations hold? Each statement tells us that neither holds because at least two of the variables are equal.

Bunuel · Answer

Official Solution:

If  x ,  y  and  z  are positive integers, is  (x - y)(y - z)(x - z) > 0 ? 
 
(1)  x^2 + yz = xy + xz 
 
To obtain information about the factors in the question ( (x - y) ,  (y - z) , and  (x - z) ), we can manipulate the given equation:
 
Rearrange:  x^2 + yz - xy - xz = 0 ;
 
Group:  (x^2 - xy) + (yz - xz) = 0 .
 
Factor out  x  and  -z , respectively:  x(x - y) - z(x - y) = 0  
 
Factor out  x-y :  (x - y)(x - z) = 0  
 
This implies that either  (x - y)  or  (x - z)  is 0, resulting in  (x - y)(y - z)(x - z)  being equal to 0, which answers the question with a NO. 
 
Sufficient. 
 
(2)  xy - y^2 = xz - yz 
 
Again, we manipulate the given equation to obtain information about the factors in the question ( (x - y) ,  (y - z) , and  (x - z) ):
 
Rearrange:  xy - y^2 - xz + yz=0 ;
 
Factor out  y  and  -z , respectively:  y(x - y) - z(x - y) = 0  
 
Factor out  x-y :  (x - y)(y - z) = 0  
 
This implies that either  (x - y)  or  (y - z)  is 0, resulting in  (x - y)(y - z)(x - z)  being equal to 0, which answers the question with a NO. 
 
Sufficient.

Answer: D

If x, y and z are positive integers, is (x - y)(y - z)(x - z) > 0?

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