Is x^3 y ? (1) x^(1/3) y (2) x^2 y

Question

Is  x^3 > y ?

(1)  \sqrt {x} > y

(2)  x^2 > y

JerryAtDreamScore · Answer

Breaking Down the Info:

We are allowed to cube both sides as that does not affect the signs of either side.

Statement 1 Alone:

Cube both sides to get  x > y^3 . This is not enough to tell whether  x^3 > y . Insufficient.

Statement 2 Alone:

We don't know if x or y is positive or negative from this. Insufficient.

Both Statements Combined:

If x is positive, we can clearly have  x^3 > y .

If x is negative, then we can have  x > y^3 , where x and y are both negative. E.g.  (-2) > (-2)^3 , this results in  x^3 < y , so combined it is still insufficient.

Answer: E

ND104 · Answer

(I)  \sqrt {x} > y  , 
case1- if x=1/8 and y=1/9,  \sqrt {1/8} > (1/9) , is  (1/8)^3 > (1/9)  -NO
case2- if x=8 and y=0,  \sqrt {8} > (0) , is  (8)^3 > (0)  -YES
statement 1 alone not sufficient

(II)  x^2 > y ,
if x=-1/2 and y=0,  (-1/2)^2 > 0 , is  (-1/2)^3 > 0  -NO
if x=8 and y=0,  (8)^2 > 0 , is  (8)^3 > (0)  -YES
statement 2 alone not sufficient

Both Togather-  \sqrt {x} > y  and  x^2 > y 
negative x not possible as sqrt {x} needs to be 0 or positive, though Y negative is possible.
If x is positive and y is negative then cube root or square of positive number x will be positive, and always be grater than negative number y, resulting  x^3 > y  YES
but if x=1/8 and y=1/65 then statement1- (1/2) >(1/65) and statement2 - (1/64)>(1/65) but  x^3 > y  NO

ANSWER-   E

Bunuel ,  KarishmaB ,  IanStewart ,  Archit3110 
can anyone tell me, is there any better way to check if both statement togather sufficient or not? algebrically or otherwise?

Archit3110 · Answer

target is x^3 > y best way is to use plug in numbers -1y as we get both yes and no to target ND104

IanStewart · Answer

When information is not sufficient, there may not be any algebraic or conceptual way to prove that, so you should quickly fall back on testing scenarios when you don't see any alternative. I think you may have made that more challenging than it needs to be, though -- I wasn't able to follow how you reached the conclusion I highlighted above, that x cannot be negative, but that's not true here (we can't take square roots or any other even root of a negative number, but we can take cube roots or any other odd root of a negative number).

So when I use both Statements here, I notice that Statement 2 will be true no matter what if y is negative, because x^2 is positive (or zero). So if I make y negative, I don't need to even think about Statement 2, because it will be true. I only need to think about Statement 1. And because we want numbers that we can easily cube root, and that change when we cube root them (we want to avoid -1, 0 or 1 here), I'll make y = -8. Then we just want to quickly try to get a 'yes' and a 'no' answer to the question "Is x^3 > y?"

• If we make y = -8 and x = 1,000,000, then both Statements are true, and the answer to the question is 'yes'

• if we make y = -8 and x = -8, then both Statements are true (because the cube root of -8 is -2, which is greater than y here), and the answer to the question is 'no'

So the answer is E. I'd only test fractions as a last resort, but in certain inequality questions that resemble this one (where you are comparing powers of unknowns), testing fractions is genuinely necessary.

bumpbot · Answer

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Is x^3 > y ? (1) x^(1/3) > y (2) x^2 > y

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