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Inequality :S

Expert replies
by anantbhatia » Sun Oct 03, 2010 12:56 pm
Is y - x > 1/(x-y) ?

(1) ┃x - y┃ > 1

(2) y > x


[spoiler]OA: B Grockit[/spoiler]

my soln:

same as:-

is ((x-y)^2+1)/(x-y) <0

with (1), I get a firm No.

with 2, I get a firm Yes. So I chose D

What was my mistake?
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Source: — Data Sufficiency |

by Rahul@gurome » Sun Oct 03, 2010 5:47 pm
anantbhatia wrote:Is y - x > 1/(x-y) ?

(1) ┃x - y┃ > 1

(2) y > x


[spoiler]OA: B Grockit[/spoiler]

my soln:

same as:-

is ((x-y)^2+1)/(x-y) <0

with (1), I get a firm No.

with 2, I get a firm Yes. So I chose D

What was my mistake?
1st statement is not appearing fine. Can you please check?
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
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by diebeatsthegmat » Sun Oct 03, 2010 5:51 pm
anantbhatia wrote:Is y - x > 1/(x-y) ?

(1) ┃x - y┃ > 1

(2) y > x


[spoiler]OA: B Grockit[/spoiler]

my soln:

same as:-

is ((x-y)^2+1)/(x-y) <0

with (1), I get a firm No.

with 2, I get a firm Yes. So I chose D

What was my mistake?
i dont understand why the answer is B. i think it should be A because
y-x>1/x-y <=> x-y<-1/(x+y) <=> x-y+1/x-y <=> ((x-y)^2+1)/x-y
since x-y>1 and x-y)^2 is always positive
thus A is sufficient
and B cant
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by anantbhatia » Sun Oct 03, 2010 8:47 pm
Is y - x > 1/(x-y) ?

(1) x-y > 1

(2) y > x

Reposted the question.

@Rahul: Please help me with the generic and quick approach to the inequalities in DS. I feel like I am struggling while giving mocks.
Join the discussion

by goyalsau » Sun Oct 03, 2010 11:16 pm
anantbhatia wrote:Is y - x > 1/(x-y) ?

(1) x-y > 1

(2) y > x

Reposted the question.

@Rahul: Please help me with the generic and quick approach to the inequalities in DS. I feel like I am struggling while giving mocks.
I think it should be A,

y - x > 1/(x-y)

(y-x) (x-y) > 1
xy - y^2 - x^2 + xy > 1
2xy - y^2 - x^2 > 1
multiplied by -ve on both sides
- 2xy + y^2 + x^2 < - 1 ( AS i know equality sign flips when we multiply by -ve )

( x - y ) ^ 2 < - 1 or ( y - x ) ^ 2 < - 1 This is question for me know ( Please correct if i am wrong because i am still not sure with -ve multiplication )

(1) x-y > 1 Then we have a positive value which is sufficient to answer

(2) y > x

let y = - ( 1/2 )
let x = - (1/3 )

or let y = 2
let x = 1

Gives us different value when we substitute it above equation so its insufficient.

HOPE this Helps..
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
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by anantbhatia » Sun Oct 03, 2010 11:33 pm
Goyal sahib, you have taken a value from RHS to the LHS without considering if it was positive or negative, which is incorrect. If you want to take it to the left, you should subtract it without affecting the inequality.
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by goyalsau » Sun Oct 03, 2010 11:53 pm
anantbhatia wrote:Goyal sahib, you have taken a value from RHS to the LHS without considering if it was positive or negative, which is incorrect. If you want to take it to the left, you should subtract it without affecting the inequality.
Thanks Dear,
as you said you struggling in inequalities, I don't know what i should call about my inequalities :P
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
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by shovan85 » Mon Oct 04, 2010 1:18 am
After going through GOYALSU and ANANTBHATIA discussion:

1: is insufficient as we cannot say whether x-y is +ve or -ve. So we ar not sure about the sign.
2: y>x
=> y - x > 0 ......(a)

so x-y < 0
so 1/(x-y) < 0 ......(b)

from (a) and(b) we can say 1/(x-y) < 0 < y - x => y-x > 1/(x-y) (Sufficient)

Now try with some values:
x=2,y=3 => y-x = 1 and x-y = -1 so 1/(x-y) = -1 (Satisfied)
x=-3,y=-2 => y-x = 1 and x-y = -1 (Satisfied)

B is Sufficient.
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by shovan85 » Mon Oct 04, 2010 1:20 am
goyalsau wrote: (2) y > x

let y = - ( 1/2 )
let x = - (1/3 )

Gives us different value when we substitute it above equation so its insufficient.

HOPE this Helps..
here y < x so u missed it ;)
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