Is x y? (1) y^2 x^2 (2) x^2 + x^3 y^2 + y^3

Question

Is  x > y ?

(1)  y^2 > x^2

(2)  x^2 + x^3 > y^2 + y^3

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Archit3110 · Answer

target is x>y
#1
  y^2 > x^2 
y = (+-2) & x (+/-1)
insufficient
#2
 x^2 + x^3 > y^2 + y^3 
x^2*(1+x)>y^2*(1+y)
two cases when -ve ; 
1: integer value of x least be 0 ,-1,-2 and y can be any value <-1/-2 or X -1/2 and y 0
case 2 : when +ve integervalue of x is 1 and y is 0 and X is -1/2 and y is 0
x>y insufficient
From 1&2 
X >y
Option C

nick1816 · Answer

Statement 1- x^2-y^2<0 (x+y)(x-y)<0 If x+y>0, xy(Yes) Insufficient Statement 2- x=-1/2 and y=0 (No) x= 1/2 and y=0 (Yes) Insufficient Combining both statements x^3>y^3 (from equation (1) and (2)) or x>y ( sufficient ) C

ramlala · Answer

A: may or man not be
B: not true if -1<x<0

Ans is C: True when both A & B

MooneeyTunes · Answer

Statement 1- x^2-y^2<0 (x+y)(x-y)<0 If x+y>0, xy(Yes) Insufficient Statement 2- x=-1/2 and y=0 (No) x= 1/2 and y=0 (Yes) Insufficient Combining both statements x^3>y^3 (from equation (1) and (2)) or x>y ( sufficient ) C Can you explain how did you get that equation on combing both?

prateekchugh · Answer

Is x>y? (1) y^2>x^2 1) x=2 and y=3 y^2>x^2 = 9>4 ; xx^2 = 9>4; x>y YES This statement is not sufficient (2) x^2+x^3>y^2+y^3 x^2(x+1)>y^2(y+1) 1) x= 3 and y=2 x^2(x+1)>y^2(y+1) = 32>12 ; x>y YES 2) x=-1/2 and y=0 x^2(x+1)>y^2(y+1) = 1/8>0 ; xx^2 & x^2+x^3>y^2+y^3 We get x^3>y^3 1) x= 3 and y=2 x^3>y^3 = 27>8 ; x>y YES 2) x=-1/2 and y=-1/3 x^3>y^3 = -1/8>-1/27 ; x>y YES Answer is C

Nevernevergiveup · Answer

Bunuel  can you please post OE for this question

Bunuel · Answer

Is  x > y ?

(1)  y^2 > x^2 ;

Take the square root:  |y| > |x| . This means that y is farther from 0 than x is. From this we cannot say whether  y > x . Not sufficient.

(2)  x^2 + x^3 > y^2 + y^3

Test numbers: if x = 1 and y = 0, then the answer is YES but if x = -2 and y = 0, then the answer is NO. Not sufficient.

(1)+(2):

Add  y^2 > x^2  and  x^2 + x^3 > y^2 + y^3  (we can safely do that since the signs are in the same direction):  y^2 + x^2 + x^3 > x^2 + y^2 + y^3 ;

Cancel  y^2 + x^2  from both sides:  x^3>y^3 ;

Take the cube root:  x > y .

Sufficient.

Answer: C.

Hope it helps.

bumpbot · Answer

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Is x > y? (1) y^2 > x^2 (2) x^2 + x^3 > y^2 + y^3

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