xy

Question

is x>y? 1) x^.5 > y 2) x^3 > y [spoiler]please confirm the answer.. my answer C yes x>y.[/spoiler]

greymatter · Answer

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

abhinav85 · Answer

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.

ssmiles08 · Answer

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)

rah_pandey · Answer

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

ankit1383 · Answer

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...........

ssmiles08 · Answer

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.

Ian Stewart · Answer

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.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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