This is the full question -Data Sufficiency

[tuichats]

Is x > y?

(1) /X > y

(2) x3 > y

item 1 is supposed to be the square root of x >y.
I have an issue with item 2,how can item 2 not be sufficient ?
Thanks!

[messi10]

Hi,

Lets take a few examples

1) When x = 2 and y = 3

x^3 = 8
In this case x^3 > y and x < y

2) When x = -1 and y = -2

x^3 = -1
In this case x^3 > y and x > y

Therefore, statement two is insufficient

If no information is given about the variables such as whether they are integers, positive, negative etc, you must consider all possible options: fractions, positives, negatives and even zero.

Regards

Sunil

[tuichats]

Thanks !

[prakhar_au]

What's the correct answer to this questions? Is it C?

[jnelson0612]

What's the correct answer to this questions? Is it C?

I think it is, because statement 1 tells you that X cannot be negative, so you can rule out that case in statement 2.

[rajat.arora]

I guess its clear that (1) and (2) are individually not sufficient.

Subtracting (2) from (1), Sq. Rt.(x) - x3>0.

Therefore, Sq. Rt. (x) > x3 and x > x6

Above implies that x lies between 0 & 1. For any number 0 & 1, if x3>y, then x has to > than y.

Regards

[JohnHarris]

I guess its clear that (1) and (2) are individually not sufficient.

Subtracting (2) from (1), Sq. Rt.(x) - x3>0.

Therefore, Sq. Rt. (x) > x3 and x > x6

Above implies that x lies between 0 & 1. For any number 0 & 1, if x3>y, then x has to > than y.

Regards

IF that is the case, one should also be able to subtract (1) from (2) to arrive at

-(Sq. Rt.(x) - x3>0) > 0

which is a contradiction to

Sq. Rt.(x) - x3>0 > 0

So subtracting that way across inequalities is not a valid operation.

[rajat.arora]

I guess its clear that (1) and (2) are individually not sufficient.

Subtracting (2) from (1), Sq. Rt.(x) - x3>0.

Therefore, Sq. Rt. (x) > x3 and x > x6

Above implies that x lies between 0 & 1. For any number 0 & 1, if x3>y, then x has to > than y.

Regards

IF that is the case, one should also be able to subtract (1) from (2) to arrive at

-(Sq. Rt.(x) - x3>0) > 0

which is a contradiction to

Sq. Rt.(x) - x3>0 > 0

So subtracting that way across inequalities is not a valid operation.

subtraction can b done with inequalities, provided they carry the same sign.. The soln can still be found if we subtract (1) from (2). This would imply x3 is greater than sq rt (x) and therefore x is greater than 1. For any number greater than 1, if sq rt(x) is greater than y, then x has to be greater than y.

[JohnHarris]

...subtraction can b done with inequalities, provided they carry the same sign..

As you point out, if x is greater than 1 you can get the answer by subtracting one way and if it is less than or equal to one you can get it by subtracting it the other way. But, since you don't know what x is, you can't do the subtracting and then declare the results for x. You have to know what x is before you can do the subtracting. You can not do subtracting across inequalities in that manner without knowing the answer before hand.

What I was trying to point out is that you are on the right track for one way to do the problem but approached it from the wrong end. Since you don't know what x is, other than non negative [by (1)], you need make your approach from the other end and break the problem into two cases such as:
(a) 0 < x < 1
and
(b) x > 1
which covers all possible values of x allowed.

In case (a) x > x^3 > y

In case (b) x > √x > y

EDIT ADDED: I am not disputing the fact that if
a > b
and
c > d
then
a + c > b + d

[tim]

Please pay close attention to this, everyone: You can NEVER validly subtract two inequalities from each other. John brought up an excellent point here in that allowing subtraction would lead to contradictory results depending on which inequality you subtracted from the other..

Anyway, I wonder where this problem came from. This actually is a classic example of the squeeze theorem from calculus: because on the domain where root(x) is defined we always have root(x) <= x <= x^3 OR root(x) >= x >= x^3, the fact that both root(x) and x^3 are greater than y means that x is as well. Please note that you NEVER need calculus to solve a GMAT problem... it just helps sometimes! John had a good solution that didn’t use calculus; calculus was just the first thing I thought of.. :)