St 1) There will be two cases.

Case 1 x=2 y=5 z=2

Case 2 x=2, y=8, z=3(yes). Therefore insufficient.

St 2) similar cases of yes and no. Condn 2 can b written as x<Z/2<y.

Case 1 x = 2, Z=6,y= 4 (no)

Case 2 x=1, z=4 y= 5 (yes).

But if we Combine 1+2 you get x<2z<y AND x<0.5z<y.

So X<Y<Z. Sufficient

So the answer is C

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