If x 1 and y 1, is x

Question

If x > 1 and y > 1, is x < y? (1) x^2 < xy + x (2) \frac{xy}{y(y − 1)} < 1

DavidTutorexamPAL · Answer

1) simplifying, we get x^2 x^2 x x< y. Sufficient! Answer B.

vishalkazone · Answer

St:1
X^2<xy+x
X( x-y-1) <0
Either x <0 or ( x-y-1) <0
X cant be less than zero ( given x>1)
So X-y-1 <0
x-y< 1
if x=1.5 and y =1.2 so answer is NO
if x=1.7 and y =1.9 so answer is YES
Not sufficient.
St:2
Given x and y positive we can cross multiply.
x/ (y-1) <1
x < (y-1)
above clearly says x < y.
Sufficient.

Answer B

fskilnik · Answer

x,y\,\, > \,\,1\, x\,\,\mathop < \limits^? \,\,y \left( 1 ight)\,\,\,{x^2} < xy + x\,\,\,\,\mathop \Leftrightarrow \limits^{x\,\, > \,\,0} \,\,\,\,x < y + 1\,\,\,\,\left\{ \matrix{ \,{ m{Take}}\,\,\left( {x,y} ight) = \left( {2,2} ight)\,\,\,\, \Rightarrow \,\,\,\left\langle {{ m{NO}}} ight angle \,\, \hfill \cr \,{ m{Take}}\,\,\left( {x,y} ight) = \left( {2,3} ight)\,\,\,\, \Rightarrow \,\,\,\left\langle {{ m{YES}}} ight angle \,\, \hfill \cr} ight. \left( 2 ight)\,\,\,{{xy} \over {y\left( {y - 1} ight)}} < 1\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{y - 1\, > \,\,0} \,\,\,\,\,{{xy} \over y} < y - 1\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{y\,\, e \,\,0} \,\,\,\,x < y - 1\,\,\mathop < \limits^{{ m{always}}} \,\,y\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{ m{YES}}} ight angle The correct answer is therefore (B). This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.

chetan2u · Answer

The most important thing is that x and y are positive,  so we can multiply or divide both sides by any of x or y without changing signs.

(1)  x^2 < xy + x.......x*x<x(y+1) 
Divide both sides by x 
 x<y+1 
If x is 2 and y is 1.5, x>y. 
If x is 2 and y is 10, x<y. 
Insufficient

(2)  \frac{xy}{y(y − 1)} < 1 
Cross multiply and divide by y. 
 xy<y(y-1).......x<(y-1)<y 
So x<y in all cases. 
Sufficient

B

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If x > 1 and y > 1, is x < y? (1) x^2 < xy + x (2) xy/(y(y − 1)) < 1

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