IMO A
s1) rut x > y
same as +-x>y^2
now y^2 is always +ve and therfore for this statement to be true x must be +ve....(a -ve no. can not be > square of any no.)
hence since x is +ve , x^3>x>y^2>y even if y is -ve
s2) not suff
when x=-2 and y=-3 x^3<y
if x=2 and y=-1 x^3>y
hence A
Wats the OA
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inequal...ds roots
Source: Beat The GMAT — Data Sufficiency |
maihuna wrote:If x and y are positive, is x^3 > y?
(1) \/x > y
(2) x > y
1. x^3 > y^6 is same as sqrt(x) > y
x^3 -y > y^6 -y
y^6 -y > 0 when y > 1: we can decide x^3 > y
y^6 - y < 0 when 0 < y < 1: here, we can't
Insufficient.
2. x > y
x^3 > y^3
x^3 -y > y^3 -y
y^3 - y > 0 when y> 1: we can decide x^3 > y
y^3 - y < 0 when 0 < y < 1: we can't say x^3 > y
Combined together.
x^3 > y^3 and x^3 > y^6
case 1: y^3 > y^6. This happens when 0 < y < 1
x^3 -y's min value: y^3 -y, which is -ve. Useless.
case 2: y^6 > y^3, this happms when y > 1
x^3 - y's min value y^6 -y, which is positive.
We can stop at case 1, n say E
true, x > y^2. how can you say that x^3 > x, unless x > 1. The question says x is +ve: it cud be (0,1) or x > 1xcusemeplz2009 wrote:IMO A
s1) rut x > y
same as +-x>y^2
now y^2 is always +ve and therfore for this statement to be true x must be +ve....(a -ve no. can not be > square of any no.)
hence since x is +ve , x^3>x>y^2>y even if y is -ve
IMO E
1> sqrt (x) > y
x > y^2
but what if we have a decimal value, for example
x = 0.1, y= 0.2
y^2 = .04 which is less than x (=.10)
so we can't be sure whether x ^3 >y
2> x > y
if x = 5, y =4 then yes we have x^3 > y
but x = 0.2, y = 0.3 , then NO because (x^3 =).008 < 0.3 (y)
so insufficient..
now combine both
x>y^2 && x> y
if x =5 , y =2 , then yes x^3 > y
but if x = 0.9, y = 0.1 then No ( .027 < .1)
1> sqrt (x) > y
x > y^2
but what if we have a decimal value, for example
x = 0.1, y= 0.2
y^2 = .04 which is less than x (=.10)
so we can't be sure whether x ^3 >y
2> x > y
if x = 5, y =4 then yes we have x^3 > y
but x = 0.2, y = 0.3 , then NO because (x^3 =).008 < 0.3 (y)
so insufficient..
now combine both
x>y^2 && x> y
if x =5 , y =2 , then yes x^3 > y
but if x = 0.9, y = 0.1 then No ( .027 < .1)
