very hard ds, pls, help

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tanviet: Legendary Member; Posts: 1404; Joined: Tue May 20, 2008 6:55 pmFollowed by: 2 members; Thanked: 18 times

by tanviet » Tue Oct 20, 2009 3:35 am

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

a, square root of x>y

b,x>y

also pls, tell me how to write the square root of x, for example square root of 9 is 3.

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Source: Beat The GMAT — Data Sufficiency | Tue Oct 20, 2009 3:35 am

xcusemeplz2009: Master | Next Rank: 500 Posts; Posts: 399; Joined: Wed Apr 15, 2009 3:48 am; Location: india; Thanked: 39 times

by xcusemeplz2009 » Tue Oct 20, 2009 6:09 am

IMO D

given; x>0 and y>0
asked x^3>y

statement 1) sqrtx>y=>+-x>y^2;since x & y is positive

x>y^2

now x^3>x(for positive value of x)
hence X^3>y
suff

Statement 2)
x>y
since X^3>X therefore X^3>y
suff.

hence D

OA pls

It does not matter how many times you get knocked down , but how many times you get up

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laptopsforlearning: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Mon Oct 19, 2009 7:48 pm

Re:

by laptopsforlearning » Tue Oct 20, 2009 6:55 am

the question, "if x and y are positive, is x^3>y ?", does it specify if it is an integer or not. Currently, the way the problem is stated, the answer would be A. If X =.2 and y=.1 then x^3 = .008 and if it x=2 and y =1 then x^3=8.

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mridula: Senior | Next Rank: 100 Posts; Posts: 44; Joined: Tue Feb 03, 2009 5:17 am

by mridula » Tue Oct 20, 2009 8:12 am

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nakul_anand: Senior | Next Rank: 100 Posts; Posts: 74; Joined: Fri Oct 09, 2009 4:20 pm; Thanked: 4 times

Answer - E

by nakul_anand » Tue Oct 20, 2009 6:05 pm

(1) sqrt(x) > y

let x = 4 and y = 1
then sqrt(x) > y

and x^3 = 64 is also > 1

but if x = 0.64 and y = 0.6
then sqrt (x) = 0.8 > 0.6

BUT x^3 = 0.26 is NOT greater than 0.6

Therefore statement 1 is not sufficient...

(2) x > y

if x = 2 and y = 1
then x > y and x^3 = 8 is greater than y = 1

but if x = 0.64 and y = 0.6
then x^3 = 0.26 is not greater than y = 0.6

therefore statement 2 is not sufficient

Statements 1 and 2 taken together

if x = 4 and y = 1
then x > y and sqrt(x) > y

and x^3 is also > y

now if x = 0.64 and y = 0.6

then x > y and sqrt (x) = 0.8 > y (0.6)

but x^3 = 0.26 < y (0.6)

Therefore options 1 and 2 taken together are also not suffient.

Hence, the answer is E

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