inequal...

maihuna wrote:If x and y are positive, is x^3 > y?
(1) (x)^1/2 > y
(2) x > y
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Either X is less than 1, or greater than equal to 1.

For X >= 1 statement 2 is enough but we don't know that x>= 1.
So it all depends on what we get for x<1.

to evaluate x < 1 lets say x = 1/4 and y = 1/8.

1/4 > 1/8 is true.
1/2 > 1/8 is also true.
but 1/8 > 1/8 is NOT true. it would have been true if the value for y we chose was 1/9. it WOULD be true.
So the givens statements are inconclusive.
My answer is 'E'



(x)^1/2 = 1/2

mridul_dave wrote:

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

Either X is less than 1, or greater than equal to 1.

For X >= 1 statement 2 is enough but we don't know that x>= 1.
So it all depends on what we get for x<1.

to evaluate x < 1 lets say x = 1/4 and y = 1/8.

1/4 > 1/8 is true.
1/2 > 1/8 is also true.
but 1/8 > 1/8 is NOT true. it would have been true if the value for y we chose was 1/9. it WOULD be true.
So the givens statements are inconclusive.
My answer is 'E' (x)^1/2 = 1/2

i m not sure if u r using both th condition simultaneously, can u please use a value that have differing values meeting both

IMO E

statement 1) rutx>y or x>y^2

for any int let x=3 and y=1; x>y^2 and x^3>(y)^6>y
but for a case of fractional value let x=1/2 and y=1/3;
x>y^2 i.e 1/2>1/9 true but
x^3=1/8 and <y=1/3

hence insuff....

statment 2)
x>y for int true
for fraction false

hence insuff...

OA pls

OA : C.

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

1. x^3 > y^6
x^3 - y > y^6 - y

When y > 1, x^3 -y > 0
When 0 < y < 1, x^3 - y > negative fraction; in this case, x^3 -y can be positive or negative.


2. x^3 > y^3
x^3 - y > y^3 -y

when y > 1, x^3 -y > 0
when 0 < y < 1, x^3-y > negative fraction; x^3 -y can be psitive or negative.


Combining toegther:

case 1. x^3 > y^3 > y^6

This is possible when 0 < y < 1.
In which case, y > y^3 > y^6

Here, y > x^3 > y^3 > y^6 or x^3 > y > y^3 > y^6 possible. Insufficiennt


case 2. x^3 > y^6 < y^3

This is possible when y > 1

x^3 > y^6 > y^3 > y

E is the answer

maihuna wrote:OA : C.

I really want to know how this answer is correct. I know my solution is not solid, but I cannot find a way to show that both statements are enuff together.
two statements dont tell you much. a sqrt of a number can be positive or negative. so it just opens up the possibilities.

Please provide the source of this problem.