Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 18 Jul 2019, 08:21

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

If x ≠ y, is y - x > 1/(x-y)?

Author Message
TAGS:

Hide Tags

Director
Joined: 12 Feb 2015
Posts: 875
If x ≠ y, is y - x > 1/(x-y)?  [#permalink]

Show Tags

15 Jun 2019, 07:12
00:00

Difficulty:

55% (hard)

Question Stats:

57% (01:52) correct 43% (01:35) wrong based on 42 sessions

HideShow timer Statistics

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

1) |$$x − y$$| > 1

2) $$y > x$$

_________________
"Please hit +1 Kudos if you like this post"

_________________
Manish

"Only I can change my life. No one can do it for me"
Manager
Joined: 20 Mar 2018
Posts: 168
Location: Ghana
Concentration: Finance, Real Estate
If x ≠ y, is y - x > 1/(x-y)?  [#permalink]

Show Tags

15 Jun 2019, 07:56
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

Posted from my mobile device
Director
Joined: 19 Oct 2018
Posts: 690
Location: India
Re: If x ≠ y, is y - x > 1/(x-y)?  [#permalink]

Show Tags

16 Jun 2019, 13:08
$$(y-x)-\frac{1}{(x-y)} >0$$
$$(y-x)+\frac{1}{(y-x)} >0$$

$$K+\frac{1}{K}>0$$, 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) $$y > x$$
Director
Joined: 20 Jul 2017
Posts: 544
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: If x ≠ y, is y - x > 1/(x-y)?  [#permalink]

Show Tags

16 Jun 2019, 14:23
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

Posted from my mobile device
CrackVerbal Quant Expert
Joined: 23 Apr 2019
Posts: 16
Re: If x ≠ y, is y - x > 1/(x-y)?  [#permalink]

Show Tags

17 Jun 2019, 09:26
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
Re: If x ≠ y, is y - x > 1/(x-y)?   [#permalink] 17 Jun 2019, 09:26
Display posts from previous: Sort by