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Is (a − 5)^2 < (b + 5)^2 ?

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Is (a − 5)^2 < (b + 5)^2 ? [#permalink]

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New post 24 Apr 2017, 04:27
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Re: Is (a − 5)^2 < (b + 5)^2 ? [#permalink]

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New post 24 Apr 2017, 06:14
Is (a − 5)^2 < (b + 5)^2 ?
Using trial and error method.
(1) a < b
Lets consider the value a=0,b=1. These values, satisfy the condition
But if we take a=-2,b=-1, the condition does not hold.
Hence, not sufficient.

(2) a + b > 5
Lets consider a=4,b=2, the condition holds good.
However, if a = 15,b = -2, the condition doesn't hold good.
Hence, not sufficient.

On combining both, since one of the number definitely has to be positive,
we can mark the solution sufficient(Option C)
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Re: Is (a − 5)^2 < (b + 5)^2 ? [#permalink]

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New post 03 Sep 2017, 10:59
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Is (a − 5)^2 < (b + 5)^2 ? [#permalink]

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New post 03 Sep 2017, 11:48
Bunuel wrote:
Is (a − 5)^2 < (b + 5)^2 ?

(1) a < b
(2) a + b > 5


Simplifying the question stem:
\((a-5)^2-(b+5)^2<0\) or \((a-5+b+5)(a-5-b-5)<0\) or \((a+b)(a-b-10)<0\).
Hence the question becomes IS \(a+b<0\) & \((a-b-10)>0\) or \((a+b)>0\) & \((a-b-10)<0\)

Statement 1: \(a<b\) or \(a-b<0\) so we can say that \(a-b-10<0\) (as it is summation of two negative terms)
but we know nothing about \(a\) & \(b\) so whether \(a+b<0\) or \(a+b>0\) can't be arrived at. For example if \(a=-3\) & \(b=-1\), then \(a+b<0\) as well as \(a-b<0\) but if \(a=1, b=2\), then \(a+b>0\) but \(a-b<0\). Hence the statement is not sufficient.

Statement 2: this implies that \(a+b>0\) but we know nothing about \((a-b)\), hence can't say that whether \(a-b-10<0\) or \(a-b-10>0\). For ex. if \(a=-1, b=7\) then \(a-b-10<0\) but if \(a=-1, b=13\) then \(a-b-10>0\). Hence insufficient

Combining 1 & 2
we know that \(a-b<0\), hence \(a-b-10<0\) and \(a+b>5\) hence \(a+b>0\). therefore \((a+b)(a-b-10)<0\). Hence sufficient

Option C

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Is (a − 5)^2 < (b + 5)^2 ?   [#permalink] 03 Sep 2017, 11:48
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