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# If x ≠ y, is y - x > 1/(x-y)?

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VP
Joined: 11 Feb 2015
Posts: 1134
If x ≠ y, is y - x > 1/(x-y)?  [#permalink]

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15 Jun 2019, 06:12
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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) $$y > x$$

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Joined: 20 Mar 2018
Posts: 512
Location: Ghana
Concentration: Finance, Real Estate
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If x ≠ y, is y - x > 1/(x-y)?  [#permalink]

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15 Jun 2019, 06:56
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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Re: If x ≠ y, is y - x > 1/(x-y)?  [#permalink]

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16 Jun 2019, 12:08
$$(y-x)-\frac{1}{(x-y)} >0$$
$$(y-x)+\frac{1}{(y-x)} >0$$

$$K+\frac{1}{K}>0$$, 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
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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16 Jun 2019, 13:23
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
Joined: 22 Apr 2019
Posts: 38
Re: If x ≠ y, is y - x > 1/(x-y)?  [#permalink]

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