is x>y?

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is x>y?

by meraGMAT1 » Wed Jan 13, 2010 7:40 pm
is x > y?

1) sqrt(x) > y

2) cube(x) > y


Here is what i think:-

1) sqrt(x) > y

Case 1: 0<x<1

Let x = 0.36 and y = 0.50

sqrt(0.36) > 0.50

=> 0.6 > 0.5 But x=0.36 and y = 0.50 --- And 0.36 > 0.50 is false

Case 2: Let x = 16 and y=3

=> and sqrt(16) > 3 -- which is true

So Case 1 is "False" and Case 2 is "true"

Hence Statement 1 is insufficient.


Statement 2:-

cube(x) > y

Case 1:

cube(2) > 7 but 2 > 7 is "false"

Case 2:

I am NOT able to find a value of x such that 0<x<1

Probably because if cube(x) > y then x>1

Is my line of reasoning correct to decide that

Answer is (B)

Is there a way to think about this problem without substituting numnbers?

Thanks a lot.

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by mehravikas » Wed Jan 13, 2010 7:52 pm
is it cuberoot(x) or x^3 ??

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by meraGMAT1 » Thu Jan 14, 2010 7:30 am
its x raised to the power 3 ( x^3)

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by sameershaik » Thu Jan 14, 2010 8:37 am
IS THE ANSWER (E) ?

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by mehravikas » Thu Jan 14, 2010 12:25 pm
answer cannot be 'B' take x = y = 2

therefore x^3 > y but x is not greater than y

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by kg » Wed Jan 20, 2010 4:47 am
I guess ans-E.

Statement1

sqx >y, or x>y^2, now choose y =1/2, 2, in first case y^2 <y, and in second case y^2>y and hence x can be > or < y
hence insufficient

Statement 2

x^3 > y, and now x >y^1/3 and choose, y=8, x > 2or y^1/3 but less than y, now choose y=0, x=0

Insufficient

Combining 2 , X>y^2, x>y^1/3 still doesnt help, hence E

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by tanviet » Wed Jan 27, 2010 1:34 am
kg wrote:I guess ans-E.

Statement1

sqx >y, or x>y^2, now choose y =1/2, 2, in first case y^2 <y, and in second case y^2>y and hence x can be > or < y
hence insufficient

Statement 2

x^3 > y, and now x >y^1/3 and choose, y=8, x > 2or y^1/3 but less than y, now choose y=0, x=0

Insufficient

Combining 2 , X>y^2, x>y^1/3 still doesnt help, hence E
we can not do like you because, maybe y<0

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by tanviet » Wed Jan 27, 2010 1:44 am
IMO C

by pluging the numbers, we can prove C is correct

pls, help

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by tanviet » Wed Jan 27, 2010 1:50 am
it is clear that x>0

if x>1, x> square roof x>y. so x>y

if 0<x<1, x>x^3>y, so x>y

C is correct.

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by dmitriyaleyev » Wed Jan 27, 2010 7:44 am
I still don't get why the answer is not D...
Please help.

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by thephoenix » Wed Jan 27, 2010 8:00 am
dmitriyaleyev wrote:I still don't get why the answer is not D...
Please help.
ans can't be D as s1) is def not suff

to prove

we are given
sqrt(x)>y

we are asked if x>y or not
let us try by pluging no's bth the cases
case1) when x>y ; prove sqrtx>y
e.g:- x=4 and y=1
x>y true
sqrtx> y--->2>1 true

case 2)
let x=0.5 and y=0.3
x>y true

but sqrtx > y...false

hencde sqrt X > Y ; but for different values results are diff

hence s1) is not suff and threfore ans can't be D

s2)
x^3>y
similarly
if x=3 and y=2
X>Y

and X^3>y

but when X=3 and Y=9

X^3>Y holds true but
X>Y does not holds true

hence insuff

Hence the ans shud be E

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by dmitriyaleyev » Wed Jan 27, 2010 8:56 am
the phoenix,

Thanks for your explanation.
I see I overlooked s2.

However,
sqrt of 0.5 --->>> appx 0.7
Therefore,

"case 2)
let x=0.5 and y=0.3
x>y true

but sqrtx > y...false "

if x=0.5 and y=0.3
sqrtx will still be more than 0.3...

or am i going crazy ? :)))