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(1) |x-y| = |y-x| .... this is a fact i.e. |3| = |-3| \(|y-x| > 1\) insufficient because the value of "y-x" can be 3 or -3, so we can't answer the question.

(2) \(y > x\) so "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.

(1) |x-y|>1 --> well this one is clearly insufficient, try x-y=2 to get a NO answer and x-y=-2 to get an YES answer (notice that both examples satisfy┃x-y|>1). Not sufficient.

(2) y>x --> this can be rewritten as y-x>0 or 0>x-y, which means that LHS=y-x=positive and RHS=1/(x-y)=negative, thus y-x>1/(x-y). Sufficient.

Re: Is y - x > 1/(x-y)? (1) x - y > 1 (2) y > x [#permalink]

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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...

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.

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

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 ?

Re: Is y - x > 1 / x-y ? (1) ┃x - y┃ > 1 (2) y > x [#permalink]

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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? _________________

Click +1 Kudos if my post helped...

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Re: Is y - x > 1 / x-y ? (1) ┃x - y┃ > 1 (2) y > x [#permalink]

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24 Jun 2013, 08:45

Hi!

I have a question.

For #1, we have (-) > 1/(+) Negative number > positive, the answer is NO

It seems to me that this simply isn't wrong, but not possible.

In another problem, we ended up with (x-1)^2<0. That's not possible, naturally, as anything squared cannot be less than zero. In that problem, which can be found here: is-x-134652.html

(x-1)^2<0 but obviously this is not possible. So even though it is incorrect, it doesn't make I.) insufficient, it's simply not counted. What am I missing here?

Thanks!

Zarrolou wrote:

Is \(y - x > \frac{1}{x-y}\) ?

(1) \(|x - y| > 1\) From this we get two cases: I)\(x-y>1\) In 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 B

This is the best approach for these questions (IMO)

A simpler solution without going too much into signs interpretation.

(y-x) > 1/(x-y) => (x-y) < 1/(x-y) (multiplying by minus both sides;basically ensuring both the variable expressions to be same) => The statement above is only possible when (x-y) is a fraction and is less than 1 or negative.

1 stmt gives both the possible cases. i.e. (x-y) is both + and -ve. 2 ensures that x-y is definitely -ve and less than 1.

In other words, (x-y) and (y-x) could be two OR -2 meaning we can't determine a single value for the inequality. INSUFFICIENT

(2) y > x

If y > x then y - x > 1/(x - y) will always have a positive # on the LHS and a fraction on the RHS. Is is true whether x and y are positive or negative. SUFFICIENT

(1) |x - y┃ > 1 (take note of x-y in the stem) x - y > 1 OR x - y < -1

We have two cases here, where x-y is either greater than 1 or less than negative 1. This means that the fraction could be either positive or negative meaning we can't be sure if the inequality y - x > 1/(x - y) holds true or not. INSUFFICIENT

(2) y > x If y>x then regardless of the signs of y and x (y, x) (y, -x) (-y, -x) y-x will be positive and 1/(x-y) will be a fraction or a negative number meaning that the inequality will always hold true. SUFFICIENT

Notice that the entire question uses (x - y) or (y - x) and not the two independently so we can easily proceed by taking y - x = a to reduce complications

Is \(a > \frac{-1}{a}\) Is \(a + \frac{1}{a} > 0\) Is \(\frac{(a^2 + 1)}{a}\) > 0 Obviously a^2 + 1 is positive so we just need to worry about the sign of 'a'. If we know whether a is positive or negative, we can answer the question.

(1) |x - y┃ > 1

|a┃> 1 implies a < -1 or a > 1. Not sufficient

(2) y > x y - x > 0 a > 0 Sufficient since we get that a > 0.

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