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If xy^2 = 1, is x > y? Updated on: 29 Nov 2023, 10:49
Originally posted by MathRevolution on 10 Jan 2016, 06:32.
Last edited by Bunuel on 29 Nov 2023, 10:49, edited 3 times in total.
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D
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If xy^2 = 1, is x > y?

(1) -1 < y < 1
(2) xy < 0
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D
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If xy^2 = 1, is x > y? 10 Jan 2016, 06:43
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MathRevolution
If xy^2 = 1, is x > y?

(1) -1 < y < 1
(2) xy < 0


*A solution is going to be uploaded in two days.
Show more


Hi,
lets see what info can we gather from the Q stem..
xy^2=1..
y^2 is positive, so x has to be positive...

we have to find if x>y?

now the statements..
(1) -1 < y < 1
y^2 will be less than 1..
xy^2=1...
x=1/y^2...
clearly x>1, and hence x>y...suff

(2) xy < 0
xy<0..
clearly one of the x or y is positive and one will be negative..
But we know that x is positive, so y will be negative..
so x>y.. suff..

ans D
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If xy^2 = 1, is x > y? 11 Jan 2016, 16:59
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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
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If xy^2 = 1, is x > y? 12 Jan 2016, 18:37
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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.
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If xy^2 = 1, is x > y? 08 Apr 2016, 10:28
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MathRevolution
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.
Show more

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...
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If xy^2 = 1, is x > y? 08 Apr 2016, 13:10
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Hi Bunuel, may I've your attention on my last comment?
Thanks...
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If xy^2 = 1, is x > y? 09 Apr 2016, 01:27
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iMyself
MathRevolution
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.

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...
Show more

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.
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If xy^2 = 1, is x > y? 09 Apr 2016, 02:18
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Bunuel
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MathRevolution
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.

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...

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.
Show more

So, how undefined is greater than 1? can you clarify it, please?? Will be appreciated. Thanks...
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If xy^2 = 1, is x > y? 09 Apr 2016, 02:26
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Bunuel

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...
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It would be undefined if x were 0 BUT we know that x is NOT 0, thus it's not undefined.
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If xy^2 = 1, is x > y? 09 Apr 2016, 03:34
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Bunuel
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Bunuel

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...

It would be undefined if x were 0 BUT we know that x is NOT 0, thus it's not undefined.
Show more
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.
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If xy^2 = 1, is x > y? 09 Apr 2016, 04:57
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Quote:
if put value of y=0, then the question stem will die, right?
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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. :)
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If xy^2 = 1, is x > y? 09 Apr 2016, 12:06
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Quote:
if put value of y=0, then the question stem will die, right?

Yes iMyself it will. ---> (x * y^2) can only be 1when x!=0 and y !=0. So what is given in question stamp we can/should only use that not anything opposite to it. :)
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Is it x! (factorial)=0 and y! (factorial)=0?, i did not get how factorial works here...
Thanks...
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If xy^2 = 1, is x > y? 09 Apr 2016, 12:33
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Quote:
if put value of y=0, then the question stem will die, right?

Yes iMyself it will. ---> (x * y^2) can only be 1when 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...
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No, thats != which means NOT EQUAL. so all i said is "X is NOT EQUAL TO ZERO" AND "Y is NOT EQUAL TO ZERO"
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If xy^2 = 1, is x > y? Updated on: 04 Jun 2019, 05:38
Originally posted by TheUltimateWinner on 09 Apr 2016, 13:15.
Last edited by TheUltimateWinner on 04 Jun 2019, 05:38, edited 1 time in total.
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Thanks, special thanks to Bunuel
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If xy^2 = 1, is x > y? 04 Jun 2019, 03:33
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MathRevolution
If xy^2 = 1, is x > y?

(1) -1 < y < 1
(2) xy < 0


*A solution is going to be uploaded in two days.
Show more

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.
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If xy^2 = 1, is x > y? 04 Jun 2019, 04:01
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If xy^2 = 1, is x > y?

(1) -1 < y < 1
(2) xy < 0


*A solution is going to be uploaded in two days.
Show more

\(x.y^2\)=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.
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If xy^2 = 1, is x > y? 04 Jun 2019, 10:53
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If xy^2 = 1, is x > y?

(1) -1 < y < 1
(2) xy < 0
Show more

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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