x>y. Is (a+2b+3c)^2 > (a-2b-3c)^2 ? 1. x = a+2b+3c 2. : DS Archive
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x>y. Is (a+2b+3c)^2 > (a-2b-3c)^2 ? 1. x = a+2b+3c 2.

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x>y. Is (a+2b+3c)^2 > (a-2b-3c)^2 ? 1. x = a+2b+3c 2. [#permalink]

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02 Aug 2005, 06:16
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x>y. Is (a+2b+3c)^2 > (a-2b-3c)^2 ?

1. x = a+2b+3c
2. y = a-2b-3c
Senior Manager
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02 Aug 2005, 07:01
x>y. Is (a+2b+3c)^2 > (a-2b-3c)^2 ?

1. x = a+2b+3c
2. y = a-2b-3c

if x and y are not both positive, this might not be correct
Director
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02 Aug 2005, 07:27
think E) is the ans. If not wrong
Manager
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02 Aug 2005, 07:35
1 not suff
2 not suff

1+2 gives
y=2a-x

y^2=4a^2+x^2-4ax

lets look at both sides of the question

is x^2>Y^2?

no; so ans is C
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02 Aug 2005, 13:16
gotoknow3 wrote:
1 not suff
2 not suff

1+2 gives
y=2a-x

y^2=4a^2+x^2-4ax

lets look at both sides of the question

is x^2>Y^2?

no; so ans is C

I guess you need to know the value of a to conclude. This value is not available, thus the answer has to be E. Wrong?
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02 Aug 2005, 17:51
C.

we donot need exact value of x and y. we can imagine any values for a, b, and c. for any values, x^2 is always greater than y^2.

thanx dan for good question.
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02 Aug 2005, 22:13
you welcome Himalaya,,

OA is E.
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03 Aug 2005, 04:54
x=1>y=-5

but x^2=1<y^2=25

if both +, x>y.

so E.
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06 Aug 2005, 22:09
People who plug in numbers need to always check negatives.

x>y does not imply x^2 > y^2 which is what this question is testing.

Because if the numbers are both negative then it isn't true !
06 Aug 2005, 22:09
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