x-y / x+y > 1 ?

[ps231]

x is not equal to y. Is x-y / x+y > 1 ?

(1) x > 0

(2) y

 

IMO : B

 

Explanation:

Multiply both sides of the inequality by x+y

x-y > x + y

 

(1) Let x = 8 and y = 7

1 > 15 - Wrong

let x = 8 and y = -7

15 > 1 - Correct

 

Condition INSUFF

 

(2) Let x = 8 and y = -7

15 > 1 - Correct

let x = -8 and y = -7

-1 > -15 - Correct

 

Hence SUFF.

 

Hence B seems to be the correct answer to me.

 

Need your views.

[getneonow]

Again you did the same mistake of cross multiplication without knowing the sign of denominator, (x+y) in this case.

 

instead bring the R.H.S(right hand side) term to L.H.S and do your simplifications.

 

i.e to verify x-y / x+y > 1

=> To verify (x-y / x+y) - 1>0

 

=>To verify [(x-y)-(x+y)] / (x+y) > 0

 

=>To verify -2y/ (x+y) > 0

 

This has to be the simplified version of original equation rather than cross multiplying to get a equation which has taken things for granted.

 

Now from here you can plug in options (1) and (2) and verify for yourself.:tup:

 

ans will be E

 

x is not equal to y. Is x-y / x+y > 1 ?

(1) x > 0

(2) y

 

IMO : B

 

Explanation:

Multiply both sides of the inequality by x+y

x-y > x + y

 

(1) Let x = 8 and y = 7

1 > 15 - Wrong

let x = 8 and y = -7

15 > 1 - Correct

 

Condition INSUFF

 

(2) Let x = 8 and y = -7

15 > 1 - Correct

let x = -8 and y = -7

-1 > -15 - Correct

 

Hence SUFF.

 

Hence B seems to be the correct answer to me.

 

Need your views.

[maychamp]

Plugging Values approach:

 

4 scenarios to consider: both +ive, Both -ive, one positive and one negative (ie x>0, y0)

 

Within each scenario try x>y and x

 

Statement 1 is insufficient because you get different results when you use any of the scenarios above

 

Statment 2 is also insufficient because you get different results when you use any of the scenarios above

 

Now Combine them i.e x>0 and y

therefore 5-(-2)/5+(-2)= 7/3

 

now pick x,y i.e. x=2, y=-5

so you get 2+5/2-5= 7/-3

 

Seeing that we get two diff results, answer is E

[ps231]

netneo,

 

thanks a lot! You were able to pinpoint my mistake :)

Hope i do not commit the same again!

[fighter_79]

Hi

The question is

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

i.e Is x>y? (using componendo-dividendo)

 

So, if x>0 and y y. So I feel, the answer should be C.

 

Please let me know the OA.

 

Regards,

[charry_008]

Answer should clearly be C. If we consider both statements then numerator would clearly be greater than denominator for all cases.

Hi

The question is

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

i.e Is x>y? (using componendo-dividendo)

 

So, if x>0 and y y. So I feel, the answer should be C.

 

Please let me know the OA.

 

Regards,

Can you please explain how you applied componendo dividendo to the above inequality.

As far as i know, componendo dividendo means

if a/b=c/d then (a+b)/(a-b)=(c+d)/(c-d)

Am I missing something here?

[trying_to_study]

Answer would be E.

 

considering both statements...

 

let x=5, y=-4, 5+4/(5-4) = 9/1 > 1

 

Now take x=3, y=-10 3+10/3-10 = negative number

 

Thus st 1 and 2 combined insuff, hence E

[shsingh]

To me it is E