Is x y? (1) x^2 y

Question

Is x > y?

(1) x^2 < y

(2) x^(1/2) >y

fskilnik · Answer

x\,\,\mathop > \limits^? \,\,y We could go straight for the (1+2) BIFURCATION , to guarantee the correct answer is (E). I will bifurcate each one separately first, for didactic reasons. \left( 1 \right)\,\,{x^2} < y\,\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {0;1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {{1 \over 2};{1 \over 3}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right. \left( 2 \right)\,\,\sqrt x > y\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {1;0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {{1 \over 4};{1 \over 3}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right. \left( {1 + 2} \right)\,\,\,{x^2} < y < \sqrt x \,\,\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\,\left( {x;y} \right) = \left( {{1 \over 4};{1 \over 4}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {{1 \over 4};{1 \over 8}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right. This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.

allan89 · Answer

St. 1: x^2y; if x<0 , y>0 e.g x=-3 ; y=1; x^2>y xy; if 0 y and xy Combining st 1 & st 2 ; root x > y > x^2 Now we know that, root x > x^2 can happen if 0 x > x^2; (remember higher powers of x reduces value when 0 x > x^2 and root x > y > x^2 ; but we still cant determine if x>y or x x > y> x^2 and root x > y> x> x^2 ; hence E Important properties; x^2y implies 0 x > root x for x>1 and x^2

Archit3110 · Answer

target is x>y #1 x^2 < y x= 1/2 & y = 1/3 we get yes to #1 & yes to target x= 0 and y =1 we get yes to #1 and no to target insufficient #2 x^(1/2) >y x=1/4 and y = 1/3 we get no to target x= 25 and y = 3 yes to target insufficient from 1 &2 value of x is +ve 0

TestPrepUnlimited · Answer

Statement 1: We know y has to be positive. If x = 2 then y > 4 and y > x must be true. On the other hand if x = 0.5, then y > 0.25 and it is possible that x > y > 0.25. Insufficient. Using my favorite inequality chain proofs, we can have x < x^2 < y or x^2 < y < x as both cases are possible. Statement 2: Use the logic from above, y < \sqrt{x} < x is possible. We may also have x < y < \sqrt{x} when x has a fractional value (0 < x < 1). Insufficient. Combined: We have x^2 < y < \sqrt{x} . This tells us x must have a fractional value (0 < x < 1). Yet x lies in between x^2 and \sqrt{x} as well, so both x and y lie in this region. Since they both have the same range any of them could be bigger, insufficient. Ans: E

Is x > y? (1) x^2 < y (2) x^(1/2) >y

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