If xy^2 = 1, is x > y?

Hello,

From the question stem we can see that x must be positive and we do not know about y. We do know that both are positives numbers.
(1) y must be a decimal number, so x must be bigger than y so the equation xy^2=1. SUFF.
(2) xy<0 , x and y have a different signal. but we know that x must be positive. Thus, y is negative. So x>y. SUFF>

Answer D

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If xy2=1, is x>y?
1) -1<y<1
2) xy<0

In the original condition, there are 2 variables(x,y) and 1 equation(xy2=1), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), in x=1/y2, it is always x>1, which is yes and sufficient.
For 2), in x=1/y2>0, x>0, y<0, which is yes and sufficient.
Therefore, the answer is D.


 For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

It would be undefined if x were 0 BUT we know that x is NOT 0, thus it's not undefined.

The option 2 is certainly wrong.
it cant be possible that xy<0 and xy^2=1,bkz at any condition both are contradicting.even you can draw a curve of y=1/x^0.5 and the entire curve will lie in 1st quadrant.

if we put the value of y=0 in the RED sentence, i'll get x=infinity. Is infinity greater than 1? if yes, how do I know that ''infinity'' is greater than 1?
Thanks...

Hi Bunuel, may I've your attention on my last comment?
Thanks...

First of all we are given in the stem that xy^2=1, which means that neither x nor y can be 0. Next, 1/0 is undefined not infinity.

So, how undefined is greater than 1? can you clarify it, please?? Will be appreciated. Thanks...

statement 1 says that y can be -.5, -.60, -.62, 0, .4, .99 etc. but zero is not allowed here according to question stem xy^2=1, because if put value of y=0, then the question stem will die, right? Thanks for help.

Yes iMyself it will. ---> (x * y^2) can only be 1 when x!=0 and y !=0. So what is given in question stamp we can/should only use that not anything opposite to it.

Is it x! (factorial)=0 and y! (factorial)=0?, i did not get how factorial works here...
Thanks...

No, thats != which means NOT EQUAL. so all i said is "X is NOT EQUAL TO ZERO" AND "Y is NOT EQUAL TO ZERO"

\(x.y^2\)=1
=> \(x=\frac{1}{y^2}\)
Therefore x is positive (as positive/ positive is positive, square of any number is always positive)

(1) -1<y<1
if -1<y<0, then y is negative. x is positive, therefore x>y. Sufficient.

(2) xy < 0, therefore
As x is +ve, therefore Y -ve. Multiple of +ve and -ve is -ve.
Therefore x>y. Sufficient.

Answer is D.

Let's do it with case testing

Question stem: xy^2 = 1. That's what you know so far. What you want to know, is whether x is bigger than y.

Statement 1: This limits the values that y could have. Try out some values for y, using both positive and negative numbers.

y = -0.5

xy^2 = 1
x(-0.5)^2 = 1
0.25x = 1
x = 4

x IS bigger than y, so the answer is "yes"


y = 0

Oops, we can't actually test this case, because it goes against the info in the question stem. The question stem says that xy^2 = 1, but that can't be true if y = 0. So, y must not equal 0, and we can ignore this case.

y = 0.5

xy^2 = 1
x(0.5)^2 = 1
0.25x = 1
x = 4

x IS bigger than y, so the answer to the question is "yes"

It seems like we're always getting "yes" answers. Just to be sure, test some extreme cases. What if y = 0.9999999 or -0.99999999?

y = 0.999999999 (very close to 1):

x(0.999999999)^2 = 1
x(something a tiny bit smaller than 1) = 1
x = 1/(something a tiny bit smaller than 1) = a little bigger than 1

So, x is greater than y again.

y = -0.9999999 (very close to -1):

x(-0.9999999999)^2 = 1
x(something a tiny bit smaller than 1) = 1
x = 1/(something a tiny bit smaller than 1) = a little bigger than 1

x is once again greater than y.

It seems like x is always greater than y, and this statement is sufficient.

Statement 2:

xy is negative. So, either x is negative and y is positive, or x is positive and y is negative.

Try to find some cases where x is negative and y is positive, and xy^2 = 1:

-2(0.5^2) = -2(0.25) = -0.5 : that doesn't work...
-1(1^2) = -1(1) = -1: that doesn't work....

In fact, there ARE no such cases, because xy^2 will only be positive if x itself is positive.

So, we're stuck with the cases where x is positive, and y is negative. But no matter what, a positive number will always be bigger than a negative one! So, x is bigger than y, and this statement is sufficient.

Since both statements are sufficient individually, the answer is D.

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