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VP  V
Joined: 11 Feb 2015
Posts: 1131
If x ≠ y, is y - x > 1/(x-y)?  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 56% (01:48) correct 44% (01:37) wrong based on 59 sessions

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If $$x ≠ y$$, is $$y - x$$ > $$\frac{1}{(x-y)}$$ ?

1) |$$x − y$$| > 1

2) $$y > x$$

_________________
Director  D
Joined: 20 Mar 2018
Posts: 509
Location: Ghana
Concentration: Finance, Real Estate
Schools: Terry '22
If x ≠ y, is y - x > 1/(x-y)?  [#permalink]

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So it says provided (x-y) is not 0
Is y-x > 1/(x-y)? Yes/No
(St.1) |x-y| >1
now, the value on the LHS of the inequality will always be positive irrespective of what the values of x or y is
Say x= 5 ,y=2 —> |5-2| >1
Subt. in the question
Is (2-5)>1/(5-3)? NO
Again x=-5,y=2 —> |-5-2| > 1
Subt. in the question
Is (2+5)> 1/(-5-2)? YES (Not suff.)

(St.2) y>x say x= 2, y = 5
Is (5-2) > 1/(2-5)? YES
Again say x=-1/2 ,y=-1/4 the same
Any number that satisfy y>x will be YES (so suff.)
IMO B

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Veritas valet et vincet
DS Forum Moderator V
Joined: 19 Oct 2018
Posts: 1826
Location: India
Re: If x ≠ y, is y - x > 1/(x-y)?  [#permalink]

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$$(y-x)-\frac{1}{(x-y)} >0$$
$$(y-x)+\frac{1}{(y-x)} >0$$

$$K+\frac{1}{K}>0$$, when K>0
$$K+\frac{1}{K}<0$$, when K<0

Hence we deduced to our new question stem that is whether y is greater than x or not.

Statement 1- |x-y|>1
y can be greater than x or less than x
Insufficient

Statement 2- y>x
Sufficient

CAMANISHPARMAR wrote:
If $$x ≠ y$$, is $$y - x$$ > $$\frac{1}{(x-y)}$$ ?

1) |$$x − y$$| > 1

2) $$y > x$$
SVP  V
Joined: 20 Jul 2017
Posts: 1506
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: If x ≠ y, is y - x > 1/(x-y)?  [#permalink]

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CAMANISHPARMAR wrote:
If $$x ≠ y$$, is $$y - x$$ > $$\frac{1}{(x-y)}$$ ?

1) |$$x − y$$| > 1

2) $$y > x$$

Is y - x > 1/(x - y) ?
—> Is (y - x) + 1/(y - x) > 0 ?

1) |$$x − y$$| > 1
—> ly - xl > 1 [since lx - yl = ly - xl

**Remember: Solution of lxl > a is x < -a or x > a
—> y - x > 1 or y - x < -1

If y - x > 1
(y - x) + 1/(y - x) = (+ve) + (+ve) = +ve (>0)

If y - x < -1
(y - x) + 1/(y - x) = (- ve) + (-ve) = -ve (<0)

Insufficient

2) $$y > x$$
—> y - x > 0 (+ve)

(y - x) + 1/(y - x) = (+ve) + (+ve) = +ve (>0)

Sufficient

IMO Option B

Pls Hit kudos if you like the solution

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CrackVerbal Quant Expert B
Joined: 22 Apr 2019
Posts: 38
Re: If x ≠ y, is y - x > 1/(x-y)?  [#permalink]

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Hi,

While solving DS inequality questions, the best approach is to always breakdown the question stem if possible. To breakdown the question stem, there are three hygiene factors, that if followed will simplify your analysis of the question.

1. Always keep the RHS of the inequality as 0
2. Simplify the LHS to a product or division of values (product and division of terms are easier to analyze)
3. Always try and maintain even powered terms (as the sign of them will always be 0 or positive)

Let us breakdown the question stem here:

Is y - x > 1/(x - y)? Keeping the RHS as 0,

Is (y - x) - 1/(x - y) > 0

Now before we take the LCM, the smarter thing to do will be to obtain a squared term wherever the opportunity arises. Remember a square will always be 0 or positive. Taking -1 common out of the denominator x - y we get,

Is (y - x) + 1/(y - x) > 0

Taking the LCM we get,

Is ((y - x)^2 + 1)/(y - x) > 0.

For this equation to hold, both the numerator and denominator have to be of the same sign. The numerator clearly here is always going to be positive, so the question can be rephrased to

Is y - x > 0?

Statement 1 : |x - y| > 1

This implies that x - y > 1 or -(x - y) > 1.

x - y > 1 -----> y - x < -1
-(x - y) > 1 -----> y - x > 1.

So y - x can be positive or negative. Insufficient.

Statement 2 : y > x.

Implies y - x > 0. Sufficient.

Takeaway : In Inequality DS questions, always spend time on simplifying the question stem and rephrase the DS question. This will help immensely while working with the two statements and will not require you to plug in values.

Hope this helps!

CrackVerbal Quant Expert
_________________ Re: If x ≠ y, is y - x > 1/(x-y)?   [#permalink] 17 Jun 2019, 08:26

# If x ≠ y, is y - x > 1/(x-y)?  