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Bunuel
If x, y and z are positive integers, is y − x > 0?


(1) y/x = z/y

(2) z > x


chetan2u - your approach
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saurabh9gupta
Bunuel
If x, y and z are positive integers, is y − x > 0?


(1) y/x = z/y

(2) z > x


chetan2u - your approach

Hi..

Info given
a) x, y and z are positive integers. SO you can cross multiply
b) is y-x>0..... or Is y>x?

(1) y/x = z/y..
\(\frac{y}{x} =\frac{z}{y}.........y^2=xz\)..
    case I : if \(x<z\), so \(z=x+a\), where a is a positive number....
    \(y^2=x(x+a)=x^2+ax\).... now ax is positive, so \(y^2>x^2.......y>x\) as y and x are positive
    case II : if \(x>z\), so \(z=x-a\), where a is a positive number....
    \(y^2=x(x-a)=x^2-ax\).... now ax is positive, so \(y^2<x^2.......y<x\) as y and x are positive
insuff

(2) z > x
nothing about y
insuff

combined..
case I of statement 1 is left
case I : if x<z, so z=x+a, where a is a positive number....\(y^2=x(x+a)=x^2+ax\).... now ax is positive, so \(y^2>x^2.......y>x\) as y and x are positive
so y>x....y-x>0
sufficient

C
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1) y/x = z/y => y^2 = xz. that means y is the middle no, however we dont know if it is in order of x,y,z or z,y,x. NOT SUFFICIENT.

2) z>x. Dont know about y. INSUFFICIENT.

1+2 , we now see z>x, so the the integers are in order of x<y<z. Thus y-x =0 .SUFFICIENT.

Correct Answer is C
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