Data Sufficiency

Is x > y?

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

(2) x^3 > y

This one was on Manhattan CAT. the explanations to this problem is through substitution of different values but I find that approach too time consuming...

Could there be alternative algebraic way of solving this?

OA C

Thx

IMO C
I really can't give you any algebraic explanation. though I can give you an advice.
for this type of questions with > or < and there are square roots involved:
check for positive value
check for negative values and
check for fractions (really important)
1) for negative and positive value x>y
but for fraction values
lets say x=1/4, y=1/4
x^1/2=1/2>y
but x not >y
NOT SUFF
2) x=5,y=3
x^3>y and x>y
x=3,y=5
x3>y but x<5
so NOT SUFF

combine 1 and 2
x>=0 because you can't find sqrt(-ve number). you can but complex numbers are not in the syllabus:)
so you are left with two options to check for:
positive values and fractions
for fractions
if x^3>y
x>y
for positive values
x^1/2>y so
x>y^2
so x>y

thanks a lot tohellandback for you detailed explanation.

well actually I do understand the explanation with testing numbers, but I just HATE doing plugging numbers because I kind of think that takes too much of my time (maybe others are very quick at it though... I ain't).

So, if anybody has algebraic version too that would be great.

ket wrote:thanks a lot tohellandback for you detailed explanation.

well actually I do understand the explanation with testing numbers, but I just HATE doing plugging numbers because I kind of think that takes too much of my time (maybe others are very quick at it though... I ain't).

So, if anybody has algebraic version too that would be great.

ket,
plugging numbers is an easy and fast approach to solving a lot a algebraic question. it saves of time. only thing is you should know when to plug numbers. That comes with practice.
for example, take this question:
k and k^2 are two roots of the equation x^2-ax+b=0. what is b+4b^2+ab.

the quickest way to solve this is plugging numbers.

maybe this reasoning could work.

you want to prove that x > y given that x^(1/2) > y and x^3 > y. Another way to prove this is to prove that x cannot be <= y and respect statement 1 and 2.

(1) x^(1/2) > y >= x ---> x^(1/2) > x ---> 0 < x < 1
(2) x^3 > y >= x ----> x^3 > x ---> x> 1

no value fulfills both equations so x cannot be <= y and therefore x > y

mikeCoolBoy wrote:maybe this reasoning could work.

you want to prove that x > y given that x^(1/2) > y and x^3 > y. Another way to prove this is to prove that x cannot be <= y and respect statement 1 and 2.

(1) x^(1/2) > y >= x ---> x^(1/2) > x ---> 0 < x < 1
(2) x^3 > y >= x ----> x^3 > x ---> x> 1

no value fulfills both equations so x cannot be <= y and therefore x > y

Thank's A lot mikeCoolBoy ... I kind of liked your reasoning but got rather confused in the end...so... I guess tohellandback must be correct in the sense that plugging in numbers is probably the best solution for this one...

p.s and also when you wrote x^3 > x ---> x> 1 ... I guess you also needed to write that x^3 > x true in case -1<x<0 numbers too.. i.e. two groups of numbers x>1 or -1<x<0 .

ket wrote:

mikeCoolBoy wrote:maybe this reasoning could work.

you want to prove that x > y given that x^(1/2) > y and x^3 > y. Another way to prove this is to prove that x cannot be <= y and respect statement 1 and 2.

(1) x^(1/2) > y >= x ---> x^(1/2) > x ---> 0 < x < 1
(2) x^3 > y >= x ----> x^3 > x ---> x> 1

no value fulfills both equations so x cannot be <= y and therefore x > y

Thank's A lot mikeCoolBoy ... I kind of liked your reasoning but got rather confused in the end...so... I guess tohellandback must be correct in the sense that plugging in numbers is probably the best solution for this one...

p.s and also when you wrote x^3 > x ---> x> 1 ... I guess you also needed to write that x^3 > x true in case -1<x<0 numbers too.. i.e. two groups of numbers x>1 or -1<x<0 .

I'm sorry you didn't understand my reasoning. Regarding your question I assume that x > 0 since statement 1 says sqrt(x) > y and in GMAT you only consider real numbers. I also think that plugging numbers is the best solution. Maybe someone else can provide another simpler algebraic solution

combine 1 and 2
x>=0 because you can't find sqrt(-ve number). you can but complex numbers are not in the syllabus:)
so you are left with two options to check for:
positive values and fractions
for fractions
if x^3>y
x>y
for positive values
x^1/2>y so
x>y^2
so x>y

I dont get this. What numbers have you used to test the conditions together? Please Elaborate.

I understand that for fractions X^3>y holds true but not necessarily for positive numbers. But I dont how u have combined the two statements here. Can you please explain.

Thanks

muna_m wrote:

combine 1 and 2
x>=0 because you can't find sqrt(-ve number). you can but complex numbers are not in the syllabus:)
so you are left with two options to check for:
positive values and fractions
for fractions
if x^3>y
x>y
for positive values
x^1/2>y so
x>y^2
so x>y

I dont get this. What numbers have you used to test the conditions together? Please Elaborate.

I understand that for fractions X^3>y holds true but not necessarily for positive numbers. But I dont how u have combined the two statements here. Can you please explain.

Thanks

Muna, lemme explain

as I said check for positive(i mean >1) values, check for negative values, check for fractions

now when we combine 1 and 2
check for >1 values
so x^1/2>y and x^3>y
x must be greater than Y

we don't need to check for negative values here because x^1/2 is not a real number if x <0
so lets check for fractions
if x^3>y,
x is always greater than y
because cube of a fraction will always be smaller than a fractional number

Hope I am clear.