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If x and y are integers and xy ≠ 0, is x - y > 0 ?

1 - x/y <1/2

2- sqr X^2 = X and sqr Y^2 = Y

Why I alone isn't sufficient , since x/y <1/2 => x<y/2 .. As X is even less than half of Y so X can't be greater than Y

Question: is \(x>y\)?

(1) \(\frac{x}{y}<\frac{1}{2}\) --> if both \(x\) and \(y\) are positive (for example 1 and 3 respectively) then \(x<y\) and the answer to the question is NO but if \(x\) and \(y\) are both negative (for example -1 and -3 respectively) then \(x>y\) and the answer to the question is YES. Not sufficient.

(2) \(\sqrt{x^2}=x\) and \(\sqrt{y^2}=y\) --> both \(x\) and \(y\) are positive (as square root function can not give negative result and we know that neither of unknown is zero), but we don't know whether \(x>y\). Not sufficient.

(1)+(2) From (2) both \(x\) and \(y\) are positive so from (1) \(x<y\) and the answer to the question is NO. Sufficient.

Answer: C.

As for your solution: red part is not correct. You can not multiply inequality by y as you don't know the sign of it.

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

Re: If x and y are integers and xy 0, is x - y > 0 ? [#permalink]

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06 Feb 2014, 10:49

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Re: If x and y are integers and xy 0, is x - y > 0 ? [#permalink]

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07 Feb 2014, 07:17

x and y are non-zero integers. Is x > y? 1. x/y < 1/2 -> x = 1 , y = 8, NO. x = -1, y = -8, YES. 2. |x| = x and |y| = y; x and y are both positive. but x > y and x < y possible. Remember Sqrt(x^2) = |x| and NOT x.

Combining, x and y are positive and x < y/2 => x < y.

Re: If x and y are integers and xy 0, is x - y > 0 ? [#permalink]

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10 Apr 2014, 02:43

shrive555 wrote:

If x and y are integers and xy ≠ 0, is x - y > 0 ?

(1) x/y < 1/2 (2) \(\sqrt{x^2}=x\)

Why I alone isn't sufficient , since x/y <1/2 => x<y/2 .. As X is even less than half of Y so X can't be greater than Y

Please note some how statement 2 has been edited , part of statement 2 is missing, \(sqrt (y^2) = y\), should be there. The question as given now , gives answer as E.

Hope concerned persons will look into this. Thank you.
_________________

Re: If x and y are integers and xy 0, is x - y > 0 ? [#permalink]

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10 Apr 2014, 04:57

stne wrote:

shrive555 wrote:

If x and y are integers and xy ≠ 0, is x - y > 0 ?

(1) x/y < 1/2 (2) \(\sqrt{x^2}=x\)

Why I alone isn't sufficient , since x/y <1/2 => x<y/2 .. As X is even less than half of Y so X can't be greater than Y

Please note some how statement 2 has been edited , part of statement 2 is missing, \(sqrt (y^2) = y\), should be there. The question as given now , gives answer as E.

Hope concerned persons will look into this. Thank you.

Hi Stne,

We do not really need \(sqrt (y^2) = y\) because if \(\sqrt{x^2}=x\) is true, x is positive and as x/y < 1/2 => 2x < y Hence as y is greater than 2x, it has to be positive.

Re: If x and y are integers and xy 0, is x - y > 0 ? [#permalink]

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10 Apr 2014, 07:05

ricsingh wrote:

stne wrote:

shrive555 wrote:

If x and y are integers and xy ≠ 0, is x - y > 0 ?

(1) x/y < 1/2 (2) \(\sqrt{x^2}=x\)

Why I alone isn't sufficient , since x/y <1/2 => x<y/2 .. As X is even less than half of Y so X can't be greater than Y

Please note some how statement 2 has been edited , part of statement 2 is missing, \(sqrt (y^2) = y\), should be there. The question as given now , gives answer as E.

Hope concerned persons will look into this. Thank you.

Hi Stne,

We do not really need \(sqrt (y^2) = y\) because if \(\sqrt{x^2}=x\) is true, x is positive and as x/y < 1/2 => 2x < y Hence as y is greater than 2x, it has to be positive.

*press Kudos if you like the post!

You cannot cross multiply in inequality without knowing the signs of x and y , you can cross multiply only when you know that the signs are positive if the question does not have \(sqrt (y^2) = y\) then y need not necessarily be positive

consider x=1 and y = 3 this satisfies both the statement and the answer is no consider x= 1 and y = -3 this satisfies both the statements and the answer is yes as you can see, Y has to be positive for the answer to be c . Hence question should be corrected, as in the present format, answer is E

Hope it helps , let me know if there is anything still unclear.
_________________

Re: If x and y are integers and xy 0, is x - y > 0 ? [#permalink]

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10 Apr 2014, 07:44

stne wrote:

ricsingh wrote:

shrive555 wrote:

If x and y are integers and xy ≠ 0, is x - y > 0 ?

(1) x/y < 1/2 (2) \(\sqrt{x^2}=x\)

Why I alone isn't sufficient , since x/y <1/2 => x<y/2 .. As X is even less than half of Y so X can't be greater than Y

Please note some how statement 2 has been edited , part of statement 2 is missing, \(sqrt (y^2) = y\), should be there. The question as given now , gives answer as E.

Hope concerned persons will look into this. Thank you.

Hi Stne,

We do not really need \(sqrt (y^2) = y\) because if \(\sqrt{x^2}=x\) is true, x is positive and as x/y < 1/2 => 2x < y Hence as y is greater than 2x, it has to be positive.

*press Kudos if you like the post!

You cannot cross multiply in inequality without knowing the signs of x and y , you can cross multiply only when you know that the signs are positive if the question does not have \(sqrt (y^2) = y\) then y need not necessarily be positive

consider x=1 and y = 3 this satisfies both the statement and the answer is no consider x= 1 and y = -3 this satisfies both the statements and the answer is yes as you can see, Y has to be positive for the answer to be c . Hence question should be corrected, as in the present format, answer is E

Hope it helps , let me know if there is anything still unclear.[/quote]

But, incase y is negative: x- y will always be greater than 0, given x is positive and x & y are non-zero integers.

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