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Manager
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x>y. Is (a+2b+3c)^2 > (a-2b-3c)^2 ? 1. x = a+2b+3c 2. [#permalink]
 02 Aug 2005, 07:16
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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
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Senior Manager
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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
if x and y are not both positive, this might not be correct
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Director
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think E) is the ans. If not wrong
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Manager
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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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Intern
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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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SVP
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Re: DS-integers,,, [#permalink]
02 Aug 2005, 18:51
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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Manager
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you welcome Himalaya,,
OA is E.
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Manager
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x=1>y=-5
but x^2=1<y^2=25
if both +, x>y.
so E.
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Manager
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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 !
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close
