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

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If x and y are integers and xy 0, is x - y > 0 ? [#permalink] New post 27 Oct 2010, 17:38
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If x and y are integers and xy ≠ 0, is x - y > 0 ?

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

[Reveal] Spoiler:
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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 11 Apr 2014, 02:45, edited 1 time in total.
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Re: integers x & y [#permalink] New post 27 Oct 2010, 17:52
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shrive555 wrote:
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.

Hope it helps.
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Re: integers x & y [#permalink] New post 27 Oct 2010, 18:37
....Great Point to remember !!! +2 :P
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Re: integers x & y [#permalink] New post 27 Oct 2010, 23:52
Really good one Bunuel... thanks
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Re: If x and y are integers and xy 0, is x - y > 0 ? [#permalink] New post 06 Feb 2014, 10:49
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Re: If x and y are integers and xy 0, is x - y > 0 ? [#permalink] New post 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.

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

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

For example 1- (-3) > 0

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Re: If x and y are integers and xy 0, is x - y > 0 ? [#permalink] New post 10 Apr 2014, 10:59
Hi Ricsingh,

x= 1 and y = 3 what is x-y ? its -2 so is x- y >0 answer No.
x=1 and y = -3 what is x- y ? its 4 so is x-y >0 answer yes

So what is the confusion? we have two different answers if sqrt(y^2)= y is not given.
So it is very important to mention that Y is positive.

Hope this will help, if any thing is still unclear .let me know
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Re: If x and y are integers and xy 0, is x - y > 0 ? [#permalink] New post 10 Apr 2014, 13:04
stne wrote:
Hi Ricsingh,

x= 1 and y = 3 what is x-y ? its -2 so is x- y >0 answer No.
x=1 and y = -3 what is x- y ? its 4 so is x-y >0 answer yes

So what is the confusion? we have two different answers if sqrt(y^2)= y is not given.
So it is very important to mention that Y is positive.

Hope this will help, if any thing is still unclear .let me know


Hello - I agree that we need to know Y is positive or not for this question and I think someone has corrected it as well.

Just a thought, if we know Y is positive we do not really need to know what X is to answer the question.

because when x is negative x-y > 0 , not true
when x is positive, x-y >0 , not true if x is less than y and first statement confirms that.

Letme know your thoughts on this.
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Re: If x and y are integers and xy 0, is x - y > 0 ? [#permalink] New post 11 Apr 2014, 02:44
Expert's post
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 and \sqrt{y^2}=y

[Reveal] Spoiler:
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

_______________________
Edited the second statement.
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If x and y are integers and xy 0, is x - y > 0 ?   [#permalink] 11 Apr 2014, 02:44
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