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ketkoag: Legendary Member; Posts: 876; Joined: Thu Apr 10, 2008 8:14 am; Thanked: 13 times

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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Source: Beat The GMAT — Data Sufficiency | Fri Jun 19, 2009 1:02 pm

greymatter: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sat May 30, 2009 2:28 am

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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abhinav85: Master | Next Rank: 500 Posts; Posts: 179; Joined: Mon Dec 29, 2008 4:41 am; Thanked: 2 times

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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ssmiles08: Master | Next Rank: 500 Posts; Posts: 472; Joined: Sun Mar 29, 2009 6:54 pm; Thanked: 56 times

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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rah_pandey: Master | Next Rank: 500 Posts; Posts: 122; Joined: Fri May 22, 2009 10:38 pm; Thanked: 8 times; GMAT Score:700

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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ankit1383: Master | Next Rank: 500 Posts; Posts: 159; Joined: Sun Jul 27, 2008 8:41 pm; Thanked: 3 times

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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ssmiles08: Master | Next Rank: 500 Posts; Posts: 472; Joined: Sun Mar 29, 2009 6:54 pm; Thanked: 56 times

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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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

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.

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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