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52) Which one is true

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by ern5231 » Wed May 12, 2010 2:21 am
Given that (x+y)^2<x^2, Which of the following can be true?

I xy<0 II y(2x+y)<0 III x<0

The OA says that I and II are right. But I found a case where I does not
hold. Try putting x=3 and y=-6

xy=18 is less than zero

but (x+y)^2 = 9
x^2 = 9

Hence condition (x+y)^2<x^2 fails because they are equal
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Source: — Problem Solving |

by kstv » Wed May 12, 2010 2:32 am
(x+y)² < x²
x and y have diff. sign

I) xy < 0 as xy will be -ve

(II) y ( 2x+y) < 0 or 2xy + y² < 0

(x+y)² = x²+(y²+2xy) < x²
x² is +ve so (y²+2xy) has to be < 0
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by ern5231 » Wed May 12, 2010 2:56 am
I did not get it yet.
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by sanju09 » Wed May 12, 2010 3:36 am
ern5231 wrote:I did not get it yet.
When (x + y) ² < x ², then x ²+ (y ²+ 2 x y) < x ², or (y ²+ 2 x y) < 0, or y ( 2 x + y) < 0, or II can be true.

Your example to deny II is perfect, but this is the trap in the wordings that are too crucial on all many occasions. The wording "can be true" extends more freedom to our imaginations than the wordings like "must be true" extend.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
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www.manyagroup.com
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