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

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For any two numbers X and Y, we define X * Y = X + Y XY. [#permalink]

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08 Jun 2009, 20:07
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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) < (Y * Y) if :

A. 1 > X > Y

B. X > 1 > Y

C. 1 >Y > X

D. Y > 1 > X

E. Y > X > 1
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rampuria

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08 Jun 2009, 23:10
IMO C
X*X = 2X-X^2
Y*Y= 2Y-Y^2
X*X<Y*Y => 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 { X<Y and X+Y <2}
Only C: 1>Y>X satisfy { X<Y and X+Y <2}

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Senior Manager
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08 Jun 2009, 23:15
Can you explain last line more elaborately. How does only C satisfy the rule. How do you logically arrive at C?
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rampuria

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09 Jun 2009, 00:07
(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?

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09 Jun 2009, 01:14
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: $$X^2<Y^2$$

Substitute in:

$$(\frac{Y}{2Y-1})^{2}<Y^{2}$$
$$(\frac{Y^2}{4Y^{2}-4Y+1})<Y^{2}$$
Inverse:
$$(\frac{4Y^{2}-4Y+1}{Y^2})>\frac{1}{Y^{2}}$$
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 $$Y<0$$ & $$Y<1$$
==>$$Y>1$$ or $$Y<1$$
==>$$\frac{X}{2X-1}>1$$ or $$\frac{X}{2X-1}<1$$ (from above)
since x>1/2 then 2x-1>0 so we can cross mult
==>$$X>2X-1$$ or $$X<2X-1$$
==>$$X<1$$ or $$X>1$$
==>($$Y>0$$ & $$X<1$$) or ($$Y<1$$ & $$X>1$$)
==>($$Y>0$$ & $$X<1$$) or ($$Y<1$$ & $$X>1$$)

Oops.
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Re: DS   [#permalink] 09 Jun 2009, 01:14
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