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Is y - x > 1/(x - y) ? [#permalink]
 05 Sep 2011, 20:48
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Is y - x > 1/(x - y) ? (1) |x - y┃ > 1 (2) y > x
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y-x > \frac{1}{-(y-x)}?3 > \frac{1}{-3} ... true-3 > \frac{1}{3} .... false(1) |x-y| = |y-x| .... this is a fact i.e. |3| = |-3| |y-x| > 1insufficient because the value of "y-x" can be 3 or -3, so we can't answer the question. (2) y > xso "y-x" will always be +ve. this is also a fact because the difference between a bigger term and a smaller term is always +ve. e.g. 3-2=1 (+ve) ... likewise ... -2-(-3)=1 (+ve) y-x > \frac{1}{-(y-x) ?}the above expression will always be true because LHS is always +ve and RHS is always -ve. sufficient. B is the ans.
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b y-x > 0 thus LHS > 0 and RHS < 0. hence always true. a LHS can be <> RHS for x<>y. thus B is clean.
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Re: Is y - x > 1/(x-y)? (1) x - y > 1 (2) y > x [#permalink]
13 Jan 2012, 14:34
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Re: Is y - x > 1/(x-y)? (1) x - y > 1 (2) y > x [#permalink]
17 Jan 2012, 23:38
Is y - x > 1/(x-y)?
(1) ┃x - y┃ > 1
x -y > 1
or
y-x > 1
not sufficient
(2) y > x
y -x > 0
x -y < 0
x -y < 0 => 1/(x-y) < 0
positive > negative.
Sufficient, hence B
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Re: Is y - x > 1/(x-y)? (1) x - y > 1 (2) y > x [#permalink]
18 Jan 2012, 00:39
Let us paraphrase!!: Suppose: y-x=t => x-y= -t,so the question simply asks us: Is t>1/-t ? 1)|-t|>1 Since |-t|=|t| => |t|>1 =>t>1 or t< -1 => we have 2 signs for t ,NOT sufficient! 2) y>x=> y-x>0 =>t>0 => t has only one sign! So, SUFFICIENT!! The answer is:B To answer this question we need to know only the sign of t! Because for all real numbers (including integers,fractions...) the positive number is always bigger than negative numbers and vice versa... Posted from my mobile device 
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Re: Is y - x > 1/(x-y) ? [#permalink]
13 Dec 2012, 00:29
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Archit143 wrote: Is y - x > 1/(x-y) ?
(1) ┃x - y┃ > 1
(2) y > x This wont remain a 700 level question, if you just observe the question and answer choices and manipulate. Entire question and answere choices revloves around x-y and y-x. Lets assume y-x = z question becomes: is z >1/-z or is z+1/z >0 ? So if we know z is positive then z+1/z is positive and we can ans, if we know z is negative then z+1/z is negative and we can ans. Only if we cant ans about the sign of z, we cant ans the qustion. Statement 1: |z| >1 => z <-1 , z >1 Now we dont if z is positive or negative, therefore not sufficient. Statement 2: y-x>0 => z>0 Now we know z is positive, and hence z+1/z is positive. Sufficient. Ans B it is!
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Re: Is y - x > 1/(x-y) ? [#permalink]
13 Dec 2012, 00:56
Vips0000 wrote: Archit143 wrote: Is y - x > 1/(x-y) ?
(1) ┃x - y┃ > 1
(2) y > x This wont remain a 700 level question, if you just observe the question and answer choices and manipulate. Entire question and answere choices revloves around x-y and y-x. Lets assume y-x = z question becomes: is z >1/-z or is z+1/z >0 ? So if we know z is positive then z+1/z is positive and we can ans, if we know z is negative then z+1/z is negative and we can ans. Only if we cant ans about the sign of z, we cant ans the qustion. Statement 1: |z| >1 => z <-1 , z >1 Now we dont if z is positive or negative, therefore not sufficient. Statement 2: y-x>0 => z>0 Now we know z is positive, and hence z+1/z is positive. Sufficient. Ans B it is! Hi Vipps can you just explain how did you get Z+1/Z >0
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Re: Is y - x > 1/(x-y) ? [#permalink]
13 Dec 2012, 01:10
Archit143 wrote: Vips0000 wrote: Archit143 wrote: Is y - x > 1/(x-y) ?
(1) ┃x - y┃ > 1
(2) y > x This wont remain a 700 level question, if you just observe the question and answer choices and manipulate. Entire question and answere choices revloves around x-y and y-x. Lets assume y-x = z question becomes: is z >1/-z or is z+1/z >0 ? So if we know z is positive then z+1/z is positive and we can ans, if we know z is negative then z+1/z is negative and we can ans. Only if we cant ans about the sign of z, we cant ans the qustion. Statement 1: |z| >1 => z <-1 , z >1 Now we dont if z is positive or negative, therefore not sufficient. Statement 2: y-x>0 => z>0 Now we know z is positive, and hence z+1/z is positive. Sufficient. Ans B it is! Hi Vipps can you just explain how did you get Z+1/Z >0 question is (y-x) > 1/(x-y) ? or is (y-x) -1/(x-y) >0 ? or is (y-x) + 1/(y-x) >0 ? Assuming (y-x) =z , it becomes is z+1/z >0 ? 
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Is y - x > 1 / x-y ? (1) ┃x - y┃ > 1 (2) y > x [#permalink]
21 May 2013, 22:58
Is y - x > 1 / x-y ?
(1) ┃x - y┃ > 1
(2) y > x
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Re: Is y - x > 1 / x-y ? (1) ┃x - y┃ > 1 (2) y > x [#permalink]
21 May 2013, 23:02
so My question is statement 1
X-y is positive so y - x is negative so the answer is a NO
and then second case is Y - X is positive so X - Y is negative so its a yes
statement 2 has Y - X is positive so its a definite yes
so this is the best approach to solve these problems or a numerical approach is better. How would solve such questions if its complex?
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Re: Is y - x > 1 / x-y ? (1) ┃x - y┃ > 1 (2) y > x [#permalink]
22 May 2013, 00:25
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Is y - x > \frac{1}{x-y} ? (1) |x - y| > 1From this we get two cases: I) x-y>1In this one we would get (-) > \frac{1}{(+)} Negative number > positive, the answer is NO II) y-x<-1(+) > \frac{1}{(-)} Positive number > negative, the answer is YES Not sufficient (2) y > x so y-x>0(+) > \frac{1}{(-)} Positive number > negative , the answer is YES Sufficient BThis is the best approach for these questions (IMO)
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Re: Is y - x > 1 / x-y ? (1) ┃x - y┃ > 1 (2) y > x [#permalink]
22 May 2013, 00:32
fozzzy wrote: so My question is statement 1
X-y is positive so y - x is negative so the answer is a NO
and then second case is Y - X is positive so X - Y is negative so its a yes
statement 2 has Y - X is positive so its a definite yes
so this is the best approach to solve these problems or a numerical approach is better. How would solve such questions if its complex? You are correct .... so where the problem lies..... From Stmt 1 .... You get multiple answers as Yes & No... So, Clearly, Insufficient. From Stmt 2 ..... Clearly Sufficient. Hence, B only.
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Re: Is y - x > 1 / x-y ? (1) ┃x - y┃ > 1 (2) y > x [#permalink]
22 May 2013, 03:09
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Re: Is y - x > 1 / x-y ? (1) ┃x - y┃ > 1 (2) y > x
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22 May 2013, 03:09
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