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OG 11 - Ques 139

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OG 11 - Ques 139

by montz » Sat Aug 18, 2007 11:52 am
If x <> -y, is (x-y)/(x+y) > 1?

1. x>0
2. y<0

Can't we simplify this to (x-y)>(x+y) so the question reduces to is 2y<0?
Then it can be answered by statement 2 alone.

The inequality in ques 114 (OG 11) has been simplified so why not this one?
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by givemeanid » Sat Aug 18, 2007 1:44 pm
You cannot multiply by (x+y) because you do not know whether x+y > 0 or x+y < 0.
So It Goes
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Re: OG 11 - Ques 139

by bingojohn » Mon Aug 20, 2007 6:42 am
montz wrote:If x <> -y, is (x-y)/(x+y) > 1?

1. x>0
2. y<0

Can't we simplify this to (x-y)>(x+y) so the question reduces to is 2y<0?
Then it can be answered by statement 2 alone.

The inequality in ques 114 (OG 11) has been simplified so why not this one?
Is OA [C]?

The reason being:
for (x-y)/(x+y) > 1 to be true,
(x-y) and (x+y) must be either both positive, or both negative
and numerator must be numerically greater than the denominator

|x-y| > |x+y|
implies,
y < 0 AND |x| > |y| .......... (a)
OR
x < 0 AND |y| > |x| .......... (b)

statement (1) says x > 0, which doesn't help.
statement (2) says y < 0, which doesn't help either.

together, eq (a) is satisfied, and is sufficient to answer the question.

Hence my answer is [C], sufficient together.

What is the OA?
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by montz » Mon Aug 20, 2007 9:26 am
Answer is E. Both statements together are not sufficient.

1. x>0
2. y<0

Combining both statements, equation (a) that you have derived is not satisfied.

y < 0 AND |x| > |y| .......... (a)

Suppose x = 2 and y = -3 then y < 0 but |x| < |y|
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by bingojohn » Mon Aug 20, 2007 9:40 am
montz wrote:Answer is E. Both statements together are not sufficient.

1. x>0
2. y<0

Combining both statements, equation (a) that you have derived is not satisfied.

y < 0 AND |x| > |y| .......... (a)

Suppose x = 2 and y = -3 then y < 0 but |x| < |y|
I knew I was missing something... you are right, answer should be [E].
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og ds 139

by simba12123 » Fri Oct 24, 2008 6:21 am
i simplified the question down to x>x+2y. I may be wrong but it seemed logical. THis questin baffles me. I am tryng to pick numbers here and its not making sense. My answer choice is B and I am sadly wrong. the contraint to the question explains that x does not equal negative. I boiled this down to x and y cannot be opposite signs. in picking numbers B seems like the right answer. help
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by rohangupta83 » Fri Oct 24, 2008 6:55 am
yup E
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Re: OG 11 - Ques 139

by yezz » Fri Oct 24, 2008 10:31 am
montz wrote:If x <> -y, is (x-y)/(x+y) > 1?

1. x>0
2. y<0

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

(x-y)-(x+y) / x+y > 0

-2y/x+y>0

is 2y/x+y<0

possible only if y -ve, x +ve , /y/>/x/or x -ve and y+ve, /x/>/y/

nothing is mentioned about absolute values thus E
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