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If x and y are positive, is x^3 > y?

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If x and y are positive, is x^3 > y? [#permalink] New post 24 Dec 2008, 14:04
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Set 1 Question 16

If x and y are positive, is x^3 > y?

1. x^1/2 > y.

x > y

Help
Director
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Re: DS: Question [#permalink] New post 25 Dec 2008, 06:25
I go with E.

1) we get , x > y^2. For positive integer we can get x^3>y but for decimals greater than 0 we dont get x^3>y. Not suff.
2) Same case with x>y.

What is OA?
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Re: DS: Question [#permalink] New post 27 Dec 2008, 05:52
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hbs2012 wrote:
Set 1 Question 16

If x and y are positive, is x^3 > y?

1. x^1/2 > y.

x > y

Help


from 1.

When x=4 and y=1 ---> x^1/2 = 2 > 1 and x^3 =64 >1 (x^3 > y)
When x=1/4 and y =1/3 --> x^1/2 = 1/2 > 1/3 and x^3 = 1/64 <1/3 (x^3 < y)

Hence (1) is not sufficient (x^3 can be greater than or less than y)

From (2):

x > y
when x = 2 and y =1 --> x^3 =8 > y =1

when x =1/2 and y=1/3 --> x^3= 1/8 < y=1/3

Hence (2) is not sufficient (x^3 can be greater than or less than y)

combining both (1) and (2) gives the same scenarios as explained above where x^3 can be greater than or less than y.

Hence answer is (E)
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Re: DS: Question [#permalink] New post 27 Dec 2008, 14:56
I agree that independtly both stms are notsuff .

But if we combine. Both stms are true if x is integer. Hence we have definite yes.

I think it C
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Re: DS: Question [#permalink] New post 27 Dec 2008, 19:24
plot y = x^3, y = sqrt(x), and y = x

All three intersect at x = 1, y = 1

Required area is BELOW x^3..

Now if you shade corresponding to A and B independently then you'll find that both result in 'maybe'

combined shading for A and B also results in 'maybe'

Thus, the solution is E
Re: DS: Question   [#permalink] 27 Dec 2008, 19:24
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If x and y are positive, is x^3 > y?

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