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If v ≠ 0, is |w| < |v| ???

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If v ≠ 0, is |w| < |v| ??? [#permalink]

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If v ≠ 0, is |w| < |v| ??? [#permalink]

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If v ≠ 0, is |w| < |v|?

(1) w/v < 1 --> if \(w=1\) and \(v=2\) the answer is YES but if \(w=-2\) and \(v=1\) the answer is NO. Not sufficient.

(2) w^2/v^2 < 1 --> since \(v^2>0\) then we can safely cross multiply: \(w^2<v^2\) --> \(|w|<|v|\). Sufficient.

Answer: B.
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Re: If v ≠ 0, is |w| < |v| ??? [#permalink]

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New post 05 Mar 2012, 09:23
Bunuel wrote:
If v ≠ 0, is |w| < |v|?

(1) w/v < 1 --> if \(w=1\) and \(v=2\) the answer is YES but if \(w=-2\) and \(v=1\) the answer is NO. Not sufficient.

(2) w^2/v^2 < 1 --> since \(v^2>0\) then we can safely cross multiply: \(w^2<v^2\) --> \(|w|<|v|\). Sufficient.

Answer: B.


I hope you are not the only user on this board, Sir.

Clear as usual. Many Thanks
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Re: If v ≠ 0, is |w| < |v| ??? [#permalink]

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New post 05 Mar 2012, 19:11
carcass wrote:
Bunuel wrote:
If v ≠ 0, is |w| < |v|?

(1) w/v < 1 --> if \(w=1\) and \(v=2\) the answer is YES but if \(w=-2\) and \(v=1\) the answer is NO. Not sufficient.

(2) w^2/v^2 < 1 --> since \(v^2>0\) then we can safely cross multiply: \(w^2<v^2\) --> \(|w|<|v|\). Sufficient.

Answer: B.


I hope you are not the only user on this board, Sir.

Clear as usual. Many Thanks


Bunuel is indeed a legend. I think we need to clone him.
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Re: If v ≠ 0, is |w| < |v| ??? [#permalink]

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New post 19 Feb 2017, 05:39
IN OPTION TWO V^2>w^2 E.G 0.9>0.16 BUT THEIR UNDER ROOT IS SIMPLY OPPOSITE V<W I.E 0.3<0.4 KINDLY CLARIFY
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Re: If v ≠ 0, is |w| < |v| ??? [#permalink]

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New post 19 Feb 2017, 08:56
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Re: If v ≠ 0, is |w| < |v| ??? [#permalink]

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New post 19 Feb 2017, 10:02
|w| < |v| and w^2 < v^2 are one and the same thing. Both can be used interchangeably so clearly 2 is sufficient.

1 will be sufficient only if both v and w are both positive or both negative. If they have opposite signs then there absolute value (distance from 0 ) can not be compared.

So Answer will be B .
Re: If v ≠ 0, is |w| < |v| ???   [#permalink] 19 Feb 2017, 10:02
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