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nikhilagrawal: Senior | Next Rank: 100 Posts; Posts: 81; Joined: Thu Jun 12, 2008 1:57 pm; Thanked: 1 times

by nikhilagrawal » Sat Oct 04, 2008 2:57 pm

Is x > y?

(1) x^(1/2)> y

(2) x^3 > y

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Source: Beat The GMAT — Data Sufficiency | Sat Oct 04, 2008 2:57 pm

sethids: Master | Next Rank: 500 Posts; Posts: 135; Joined: Mon Jun 23, 2008 3:56 am; Thanked: 7 times

by sethids » Sat Oct 04, 2008 3:18 pm

Stmt1:
x^(1/2) > y
Square both sides:
x > y^2 implies x > y. Sufficient.

Stmt2:
x^3 > y.
suppose x = 2, y = 4
x^3 = 8 is > y but x < y.
Suppose x = 2, y = 1
x^3 = 8 is > y and x > y. Insufficient.

Hence answer should be A.

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raju232007: Master | Next Rank: 500 Posts; Posts: 133; Joined: Sat Dec 29, 2007 2:43 am; Thanked: 12 times

by raju232007 » Sat Oct 04, 2008 10:51 pm

I think the ans should be C....

In GMAT the square root of a number always refers to the positive root...So in this case we have to assume x to be a positive number

statement 1:
sqrt(x)>y
Let x=4 and y=1
then sqrt(4)>1
Let x=1 and y=4
then sqrt(1)<4
statement 1 is not sufficient

statement 2:
x^3>y
let x=2 and y=1 ......x^3>y
let x=1 and y=2 ......x^3<y
statement 2 is not sufficient

Combining both statements 1 and 2
x=2 y=1 we get x>y
x=1 y=2 we get x<y
x=1/4 y=1 we get x<y
x=16 y=-2 we get >y

As a general rule if both square root and cube of a number is greater than the other number,the number is always greater than the other number.

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Morgoth: Master | Next Rank: 500 Posts; Posts: 316; Joined: Mon Sep 22, 2008 12:04 amFollowed by: 1 member; Thanked: 36 times

by Morgoth » Sat Oct 04, 2008 11:26 pm

raju232007 wrote:I think the ans should be C....

In GMAT the square root of a number always refers to the positive root...So in this case we have to assume x to be a positive number

statement 1:
sqrt(x)>y
Let x=4 and y=1
then sqrt(4)>1
Let x=1 and y=4
then sqrt(1)<4
statement 1 is not sufficient

You cant take x=1 & y=4

because these values do not hold true for statement I

You dont have to prove the statement I, it is given as a fact.

statement 2:
x^3>y
let x=2 and y=1 ......x^3>y
let x=1 and y=2 ......x^3<y
statement 2 is not sufficient

You cant take x=1 & y=2

because these values do not hold true for statement 2

You dont have to prove the statement 2, it is given as a fact.

As a general rule if both square root and cube of a number is greater than the other number,the number is always greater than the other number.

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4meonly: Legendary Member; Posts: 891; Joined: Sat Aug 16, 2008 4:21 amFollowed by: 1 member; Thanked: 27 times; GMAT Score:660(

by 4meonly » Sat Oct 04, 2008 11:48 pm

for me answer is C
combining 1 and 2 we get that x is an integer, sqrt(x) and x^3 are greater than y, so x>y

OA?

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raju232007: Master | Next Rank: 500 Posts; Posts: 133; Joined: Sat Dec 29, 2007 2:43 am; Thanked: 12 times

by raju232007 » Sun Oct 05, 2008 1:59 am

OA is C

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sethids: Master | Next Rank: 500 Posts; Posts: 135; Joined: Mon Jun 23, 2008 3:56 am; Thanked: 7 times

by sethids » Sun Oct 05, 2008 2:53 am

I agree with C now.
I didn't consider sqrt(x) could be a fraction and then what I posted earlier regarding Stmt1 would not hold true.

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