For any two numbers X and Y, we define X * Y = X + Y XY.

Question

For any two numbers X and Y, we define X * Y = X + Y – XY. Suppose that both X and Y are larger than 0.5. Then (X * X) X > Y

B. X > 1 > Y

C. 1 >Y > X

D. Y > 1 > X

E. Y > X > 1

ngoctraiden1905 · Answer

IMO C X*X = 2X-X^2 Y*Y= 2Y-Y^2 X*X 2X-X^2<2Y-Y^2 =>2(X-Y)<(X-Y) (X+Y) => (X-Y) (X+Y-2) >0 => {X>Y and X+Y>2} or { XY>X satisfy { X

rampuria · Answer

Can you explain last line more elaborately. How does only C satisfy the rule. How do you logically arrive at C?

Sunchaser20 · Answer

(X * X) < (Y * Y) when |X|<|Y|, plus given x>0.5 and y>0.5 => A and B out
Transform equation: X * Y = X + Y – XY to 2XY = X+Y
E: if y>x>1, then 2xy will always be bigger than x+y
C: if 1>y>x>0.5, then 2xy will always be smaller than x+y
Only D left: for example y=3 and x=0.6

Solved through picking numbers, quite time consuming. Maybe a faster way?

Hades · Answer

Not even close to being an official question, so not sweating much over.

XY = X+Y - XY 
 2XY = X+Y 
 2XY-Y=X 
 Y(2X-1)=X 
 Y=\frac{X}{2X-1} 
(likewise  X=\frac{Y}{2Y-1} )

Condition:  X^2\frac{1}{Y^{2}} 
Multiply by  Y^{2}  since >0
 4Y^{2}-4Y+1>1 
 4Y^{2}-4Y>0 
 Y^{2}-Y>0 
 Y(Y-1)>0 
==>  Y>0  &  Y>1  or  Y Y>1  or  Y \frac{X}{2X-1}>1  or  \frac{X}{2X-1}1/2 then 2x-1>0 so we can cross mult
==> X>2X-1  or  X X1  
==>( Y>0  &  X1 ) 
==>( Y>0  &  X1 )

Oops.

For any two numbers X and Y, we define X * Y = X + Y XY.

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