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Is X^3>y^2

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by thephoenix » Fri Mar 12, 2010 12:33 am
Is X^3>y^2----->x and y can be a int or fraction and for bth the cases we get two possibilities provided we know that how x is related to y
so our task is to find realtion b/n x and y and whether x and y are int or fraction????
if know the ans to abve we can solve it

1). y=x^2
---->x^3>x^4(substituting y in original)
if x is int then no equality does not holds true
and if x is a fraction then yes it holds true
two possibilities hence insuff...
2). x>x^2
---->x is a fraction---->x^3<x^2 but we dnt know anything abt y so inconclusive

combining yes we have an ans that yes x^3 is > x^4---->x^3>y^2
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by sanju09 » Fri Mar 12, 2010 5:31 am
gmatmachoman wrote:Is X^3>y^2

1). y=x^2

2). x>x^2
Statement (1) tells us that y ≥ 0, hence we have to, at first sight, confirm, is x^3 > 0? OR, is x > 0? And we can't. Insufficient

Statement (2) tells us that 0 < x < 1, which, by itself, cannot answer "Is x^3 > y^2?" Insufficient

Let's combine the two statements and take a suitable value for x.

When x = ½, y = ¼, and x^3 = 1/8 with y^2 = 1/16; such that

1/8 > 1/16, or [spoiler]YES, x^3 > y^2

C[/spoiler]
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by November Rain » Sat Mar 20, 2010 8:26 am
Hi

I think it's E

According with what was said previously Statement 1 ad 2 alone are not enough.


And putting them together won't be enough either, as we wll get that X>Y

if X = 3, and Y = 2, the answer is Yes.

But if X = (1/11) and Y = (1/12) the answer is No
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by gmatmachoman » Sat Mar 20, 2010 8:40 am
November Rain wrote:Hi

I think it's E

According with what was said previously Statement 1 ad 2 alone are not enough.


And putting them together won't be enough either, as we wll get that X>Y

if X = 3, and Y = 2, the answer is Yes.

But if X = (1/11) and Y = (1/12) the answer is No
@Rain
You are not Correct.

Plz chk with ur calculation for X= (1/11). You get Y value from X and dont use a seperate Y value all together. Use st2 to get value of Y in st 1.
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by rockeyb » Sat Mar 20, 2010 9:46 pm
Is X^3>y^2

1). y=x^2

2). x>x^2


This is how I would solve this :

IS x^3 > y^2 ?

Statement 1 : y = x^2

Rephrase : x = root of y ; root of y can not be -ve . So x has to be +ve .

But we dont know if x is integer or a fraction .

So only thing we know is x is + ve .

Not Sufficient .


Statement 2 : x > x^2

Rephrase : 0 < x < 1 that is x lies between 0 and 1 . But we don't know any thing about Y not sufficient .

Combine 1 and 2 we get

y = x^2 -------- (1)


and x > x^2


from (1) x > y . And (0< x) We now know the relation between x and y . Question can be answered.

Sufficient . C
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by neoreaves » Sun Mar 21, 2010 12:01 am
Let me be brave enough to disagree with you all. In my opinion it should be A.
Is X^3>y^2

1). y=x^2

2). x>x^2
1) Thus, based on y=x^2

y is positive
and y^2 = x^4

Thus if x is positive y^2 > x^3
if x is negative --> we already know y is positive so again y^2 > x^3
Thus Sufficient

2) We only have information about x. 0<x<1. This is not enough. Insufficient.
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by rockeyb » Sun Mar 21, 2010 12:29 am
neoreaves wrote:Let me be brave enough to disagree with you all. In my opinion it should be A.
Is X^3>y^2

1). y=x^2

2). x>x^2
1) Thus, based on y=x^2

y is positive
and y^2 = x^4

Thus if x is positive y^2 > x^3
if x is negative --> we already know y is positive so again y^2 > x^3
Thus Sufficient

2) We only have information about x. 0<x<1. This is not enough. Insufficient.

ARE you sure x and y are integers ?

Ex : If x - 1/2
x^2 = 1/4 and x^3 = 1/8

so y = 1/4 and y^2 = 1/16

IS x^3 > y^2

IS 1/8 > 1/16 yes .

Now lets put x = 5/2 .

x^2 = 25/4 and x^3 = 125/8

x^2 = y = 25/4

y^2 = 625/16

IS 125/8 > 625/ 16 ?

Still do you think the answer is A ?
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by neoreaves » Sun Mar 21, 2010 1:06 am
You are right ...it can't be A then
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by Stuart@KaplanGMAT » Sun Mar 21, 2010 1:53 pm
gmatmachoman wrote:Is X^3>y^2

1). y=x^2

2). x>x^2
Step 1: Analyze the Problem.

We see "is", we think "this is a yes/no question; a definitely yes is sufficient, a definite no is sufficient, a maybe/unsure is insufficient".

We want the relationship between x^3 and y^2, so we need info about both of them.

What don't we know? Everything, including what kinds of numbers x and y could be (positive/negative/0, integers/fractions, etc...).

Step 2: Evaluate the Statements.

(2) only info about x, so clearly insufficient: eliminate B and D.

(1) y = x^2

If y=x=1, then we ask: Is 1>1? NO.

If x=1/2 and y=1/4, then we ask: Is 1/8 > 1/16? YES.

We can get both a yes and a no, insufficient: eliminate A.

Combining the statements:

For x > x^2, x must be a positive fraction. We've already seen that picking x=1/2 gives us a "yes" answer. At this point we can try a couple of other fractions to see if it makes a difference, or solve algebraically to prove.

Algebraically:

(1) tells us that y = x^2, so we can rephrase the question as:

Is x^3 > (x^2)^2?

or

Is x^3 > x^4?

An important property of positive fractions is that they get smaller as you raise the exponent; accordingly, since x is a positive fraction, we'll always get a yes answer - sufficient: choose C.
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by tanviet » Sun Mar 21, 2010 8:42 pm
because x>x^2

x-x^2>0

x(1-x)>0

0<x<1

this is why x must be positive fraction
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