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700+ Inequality HELP!

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700+ Inequality HELP!

by righty123 » Fri Feb 19, 2010 5:33 pm
From Quant OG11 q.139

If x does not equal -y, is [(x-y) / (x+ y)] > 1?

a)x>0
b)y<0

I have read the ans choices and see how the ans choices fit (E). However, when I first saw the question, I manipulated the inequality to:

Does x - y > x + y (multiply both sides by x + y)

Therefore, when a & b are taken together, x - (-y) always > x + (-y) lead leading to ans (C).

Also, one can even further reduce x - y > x + y to 0 > y; which is solved by ans (B).

Any correction of my thought process is truly appreciated!
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by ajith » Fri Feb 19, 2010 9:04 pm
righty123 wrote:From Quant OG11 q.139

If x does not equal -y, is [(x-y) / (x+ y)] > 1?

a)x>0
b)y<0

I have read the ans choices and see how the ans choices fit (E). However, when I first saw the question, I manipulated the inequality to:

Does x - y > x + y (multiply both sides by x + y)

Therefore, when a & b are taken together, x - (-y) always > x + (-y) lead leading to ans (C).

Also, one can even further reduce x - y > x + y to 0 > y; which is solved by ans (B).

Any correction of my thought process is truly appreciated!
You can multiply both sides of an inequality by a positive number and keep the sign - there is no problem
Or You can multiply both sides of an inequality by a -ve number and reverse the sign

But you cannot multiply both sides of an inequality with a value ambiguous in sign'

When u multiplied with x+y you just did that, you don't know the sign of x+y
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by shashank.ism » Sun Feb 21, 2010 1:53 am
righty123 wrote:From Quant OG11 q.139

If x does not equal -y, is [(x-y) / (x+ y)] > 1?

a)x>0
b)y<0

I have read the ans choices and see how the ans choices fit (E). However, when I first saw the question, I manipulated the inequality to:

Does x - y > x + y (multiply both sides by x + y)

Therefore, when a & b are taken together, x - (-y) always > x + (-y) lead leading to ans (C).

Also, one can even further reduce x - y > x + y to 0 > y; which is solved by ans (B).

Any correction of my thought process is truly appreciated!
If x does not equal -y ,

Now for checking [(x-y) / (x+ y)] > 1
St.1 x>0 we d on't know anything abt y insuff.
St.2 y<0 we d on't know anything abt x insuff.

combined: x>0 and y<0, so x-y adds up
and x+y so x is decreased by y so overall [(x-y) / (x+ y)] > 1...suff.
Ans C

I think u are correct in ur approach ...
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by righty123 » Sun Feb 21, 2010 12:16 pm
Unfortunately, Shank, the answer according to OG is E and NOT C.
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by harsh.champ » Sun Feb 21, 2010 12:34 pm
righty123 wrote:Unfortunately, Shank, the answer according to OG is E and NOT C.
I agree with you righty123.
What shashank did was same as what ajith has cautioned against.

shashank.ism wrote:ombined: x>0 and y<0, so x-y adds up
and x+y so x is decreased by y so overall [(x-y) / (x+ y)] > 1...suff.
Agreed that x-y adds up so the value would be greater.
But for x+y since we dont know the magnitude of x and y we can't say what sign the equation will take(+ve or -ve) .

Hence,x+y can be a -ve quantity which would be less than 1.

I will show it by an example:-
Suppose x=4 , y=-2 [It follows both the statements:- x>0 and y<0]
Then,
x-y = 4-(-2) =6
x+y = 4-2 = 2
Hence, [(x-y) / (x+ y)] =3 >1

But in case,
x=2 and y=-4
Then,
x-y = 2-(-4) =6
x+y = 2-4 = -2
Hence, [(x-y) / (x+ y)] =-3 < 1

Hence we find that combining both also we can get 2 different answers.
Hence insufficient.So, E would be the answer.

Note:-In equality,for multiplication and division,always check whether it is +ve or -ve.
Last edited by harsh.champ on Sun Feb 21, 2010 12:38 pm, edited 1 time in total.
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by shashank.ism » Sun Feb 21, 2010 12:38 pm
harsh.champ wrote:
righty123 wrote:Unfortunately, Shank, the answer according to OG is E and NOT C.
I agree with you righty123.
What shashank did was same as what ajit has cautioned against.

shashank.ism wrote:ombined: x>0 and y<0, so x-y adds up
and x+y so x is decreased by y so overall [(x-y) / (x+ y)] > 1...suff.
Agreed that x-y adds up so the value would be greater.
But for x+y since we dont know the magnitude of x and y we can't say what sign the equation will take(+ve or -ve) .
Harsh I got ur point if -ve and y has magnitude greater than x then the sign of lower term will become -ve which will not satisfy the condition..
so righty123 Ans is E
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by girish3131 » Fri Feb 26, 2010 5:55 am
Guys.. i go through all yr comments.... by t hat approach ans is E ofcourse... but 1 ques...

WHY the simple way of doing the ques is not being followed here....... like x-y/x+y >1

i.e. x-y > x+y

i.e. x got cancelled out and we get y < 0 i.e. B ans.

even we don't know sign of x+y still no issue there... b'coz x-y will be always greater than x+y

Thanks!
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by ajith » Fri Feb 26, 2010 6:14 am
girish3131 wrote:Guys.. i go through all yr comments.... by t hat approach ans is E ofcourse... but 1 ques...

WHY the simple way of doing the ques is not being followed here....... like x-y/x+y >1

i.e. x-y > x+y

i.e. x got cancelled out and we get y < 0 i.e. B ans.

even we don't know sign of x+y still no issue there... b'coz x-y will be always greater than x+y

Thanks!
If x does not equal -y, is [(x-y) / (x+ y)] > 1?

a)x>0
b)y<0

This is the original question

let us take some numbers

x= -3 y = -2

x-y/x+y = -1/-5 <1


Now consider

x=3 y= -2
x-y/x+y = 5/1 >1

in both cases y<0

But the answer changes, isn't it?


On a more technical note
if (x+y) is negative
if you multiply both sides of the inequality by a -ve value the sign has to change.

x-y will not be greater than x+y - well no, precisely because x+y can be negative
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by kstv » Fri Feb 26, 2010 6:46 am
If x does not equal -y, is [(x-y) / (x+ y)] > 1?

a)x>0
b)y<0

a) & b) together
Numerator (x-y) will always be +ve

Denominator (x+y) will be +ve if x>|y|, -ve if x<|y|
taking two diff nos. x=3 and 2 correspondingly y = -2 and -3

the sign of [(x-y) / (x+ y)] will be +ve and -ve depending on the Denominator.

should this not affect the answer. Or am I missing something.
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by girish3131 » Fri Feb 26, 2010 11:47 pm
So Ajith


that means Ans is B.

N one thing .... why u r trying to solve this ques with taking values...


if u solve eq as it is as ui have solved then ans should be B


if u still convinced about ans E then plz explain it with equation... (by not taking values for X N Y)

Thanks!
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by ajith » Sat Feb 27, 2010 1:33 am
girish3131 wrote:So Ajith


that means Ans is B.

N one thing .... why u r trying to solve this ques with taking values...


if u solve eq as it is as ui have solved then ans should be B


if u still convinced about ans E then plz explain it with equation... (by not taking values for X N Y)

Thanks!

On why numbers

1. One example against a theorem can disprove it while 100 examples in favour of it cannot prove it.
2. That is the easiest method to demonstrate

On why I think answer is E

If x does not equal -y, is [(x-y) / (x+ y)] > 1?

a)x>0
b)y<0

You say answer is B, you don't provide any reasoning!
If your contention is
(x-y) / (x+ y) >1
when y<0

I disprove it by an example of x= 1 and y =-2

If your contention is
(x-y) / (x+ y) <1
when y<0

I disprove it by an example of x= 3 and y = -2

Now B is out of Question
A is also out of question by that example

C, is also out of question

All that left is E

I have tried to explain you in technical terms also what is going wrong


x-y/x+y >1

Does not mean

x-y > x+y

It depends on the sign of x+y

If (x+ y) is negative

x-y/x+y >1

=> x-y <x+y

if x+y is positive

x-y/x+y >1

=>
x-y> x+y

and y is -ve

That is the fundamental flaw in your argument. And again, going by values is the best method to disprove something if not to prove
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