Data Sufficiency

If x and y are positive, is x^3 > y?
(1) \/x > y
(2) x > y

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

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

xcusemeplz2009 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

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 > 1

I am also getting E. Please explain the answer and what is the OA?

I am lost here. Can any one give a simple explanation for choosing E ?

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)