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Interesting GMATFix Problem-45

by arora007 » Mon Oct 04, 2010 4:21 am
If x notequalto y and y notequalto -1,is [ x/(x-y) ] < [(y-x) /(2+2y) ] ?
1) x>0>y
2) y<-1
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Source: — Data Sufficiency |

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by Rahul@gurome » Mon Oct 04, 2010 4:53 am
Question is: 2x(1 + y) < -(x - y)^2 or 2x + x^2 + y^2 < 0?

(1) x > 0, y < 0
x^2 and Y^2 will always be positive as they are perfect squares and since x > 0 so 2x > 0
So, 2x + x^2 + y^2 > 0, which means answer to the main question i s"no".
So, (1) is SUFFICIENT.

(2) y < -1, so y will always be negative. x^2 and y^2 will always be positive whether x is positive or negative.
If x is positive, then 2x + x^2 + y^2 > 0
If x is negative, then also 2x + x^2 + y^2 > 0
If x = 0, then also 2x + x^2 + y^2 > 0

So, 2x + x^2 + y^2 > 0, which means answer to the main question i s"no".
So, (2) is SUFFICIENT.

[spoiler]The correct answer is (D).[/spoiler]
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by kapil403 » Mon Oct 04, 2010 8:01 am
Rahul I don't think we can generalize given statement to:
Question is: 2x(1 + y) < -(x - y)^2

if -1<y<0 E.g. -0.5, then (1+y) = +ve and sign will remain same
if y< -1 E.g. -2 , then sign will change
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by arora007 » Tue Oct 05, 2010 6:38 am
OA is C.


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by anantbhatia » Tue Oct 05, 2010 8:20 am
@Rahul: You have incorrectly multiplied the denominator on RHS to LHS. Without knowing the sign of the denominator, we cannot do so.

Simplifying,

we need to find,

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

From (1) we know that the numerator will be positive. Also (x-y) in denominator will be positive
But we cannot determine sign of (1+y)

From (2) we cannot even be sure of (x-y)'s sign.

Combining, all other things being positive, we get, 1+y < 0
A definite answer Yes to what we wanted to find. Hence (C)
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by Rahul@gurome » Tue Oct 05, 2010 5:28 pm
anantbhatia wrote:@Rahul: You have incorrectly multiplied the denominator on RHS to LHS. Without knowing the sign of the denominator, we cannot do so.

Simplifying,

we need to find,

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

From (1) we know that the numerator will be positive. Also (x-y) in denominator will be positive
But we cannot determine sign of (1+y)

From (2) we cannot even be sure of (x-y)'s sign.

Combining, all other things being positive, we get, 1+y < 0
A definite answer Yes to what we wanted to find. Hence (C)
U r right, OA is (C).
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