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x>y

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x>y

by ketkoag » Fri Jun 19, 2009 1:02 pm
is x>y?
1) x^.5 > y
2) x^3 > y

[spoiler]please confirm the answer.. my answer C yes x>y.[/spoiler]
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Re: x>y

by greymatter » Fri Jun 19, 2009 11:05 pm
ketkoag wrote:is x>y?
1) x^.5 > y
2) x^3 > y

[spoiler]please confirm the answer.. my answer C yes x>y.[/spoiler]
cosider x=2 and y =3

2^5>3

and

2^3>3

combining

2^2>3

so is x>y NO

now consider

x=3 and y =2

3^5>2
3^3>2

and
3^2>2

so is x>y Yes

hence answer should be E
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by abhinav85 » Sat Jun 20, 2009 5:41 am
i think IMO A

Becoz statement 1 says that x^.5 > y?

that means x^1/2 > y

= square root x > y!!!! sufficient.

statement 2 is insufficient.
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by ssmiles08 » Sat Jun 20, 2009 6:04 am
I got (E) as my answer.

1) x^(1/2) > y

values to test case 1.

(1/4,0) (1/4,1/4)

a. (1/4)^(1/2) > 0;
1/4 > 0 : TRUE.

b. (1/4)^(1/2) > 1/4;
1/4 > 1/4 : FALSE.

INSUFFICIENT.


2) x^3 > y

values to test case 2. (1,0) (-1/4, -1/4)

a. 1^3 > 0
1 > 0: TRUE

b. (-1/4)^3 > (1/4) = -1/64 > -1/4
(-1/4) > (-1/4): FALSE.

Together, the values of x are restricted to be positive because of case 1.

Making all x positive would result in the same scenario as case 1, which would provide contradicting values.

So i will go with (E)
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by rah_pandey » Sat Jun 20, 2009 6:18 am
Ketkoag,
have you written the entire question? does it comment on possible values x,y can take say x>0,y>0

because root(x)=> x>0
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by ankit1383 » Sat Jun 20, 2009 6:48 pm
ssmiles08 wrote:
1) x^(1/2) > y

values to test case 1.

(1/4,0) (1/4,1/4)

a. (1/4)^(1/2) > 0;
1/4 > 0 : TRUE.

b. (1/4)^(1/2) > 1/4;
1/4 > 1/4 : FALSE.

INSUFFICIENT.

Making all x positive would result in the same scenario as case 1, which would provide contradicting values.
i am still getting my answer as C......ssmiles08 can you elaborate on the bold part...........
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by ssmiles08 » Sun Jun 21, 2009 11:11 am
ankit1383 wrote:
ssmiles08 wrote:
1) x^(1/2) > y

values to test case 1.

(1/4,0) (1/4,1/4)

a. (1/4)^(1/2) > 0;
1/4 > 0 : TRUE.

b. (1/4)^(1/2) > 1/4;
1/4 > 1/4 : FALSE.

INSUFFICIENT.

Making all x positive would result in the same scenario as case 1, which would provide contradicting values.
i am still getting my answer as C......ssmiles08 can you elaborate on the bold part...........
ankit1383,

Thank you very much for catching my error. You are right. I am getting C as well. I realized that I did not fully test case 2 values on both case 1 and 2.

x has to be positive for both scenarios to work, sqrrt(x) >y and x^3 > y. x will always be > than y. Examples can be (1/4,0) (5,3) and so on.

I will go with C then.
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Re: x>y

by Ian Stewart » Sun Jun 21, 2009 12:07 pm
ketkoag wrote:is x>y?
1) x^.5 > y
2) x^3 > y

[spoiler]please confirm the answer.. my answer C yes x>y.[/spoiler]
We can assume that x is not negative; if x were negative, Statement 1 wouldn't make sense. Now:

S1 is insufficient. x could be 1/4, and y could be 1/3, for example, so x might be less than y. x could also be greater than y.

S2 is insufficient. x could be 2, and y could be 3, for example, so x might be less than y. x could also be greater than y.

S1+S2 together:

If 0 < x < 1, then x^3 < x < x^0.5. So in this case, since S2 is true, then y < x^3 < x < x^0.5, and in particular, y < x.

If 1 < x, then x^0.5 < x < x^3. So in this case, since S1 is true, y < x^0.5 < x < x^3, and in particular, y < x.

So together the statements are sufficient. C.
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