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# If mv < pv< 0, is v > 0?

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If mv < pv< 0, is v > 0? [#permalink]

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20 May 2012, 14:14
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53% (01:45) correct 47% (00:44) wrong based on 104 sessions

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If mv < pv< 0, is v > 0?

(1) m < p
(2) m < 0
[Reveal] Spoiler: OA

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20 May 2012, 18:18
Note that both mv and pv are negative.

Statement 1: m<p
=> mv<pv if v is positive (and is given to us)
=> mv>pv is v is negative

Therefore v is positive. Sufficient.

Statement 2: m<0
=> mv<0 only if v is positive (and this is given to us)

Therefore v is positive.

Sufficient.

D it is.
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Re: If mv < pv< 0, is v > 0? [#permalink]

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21 May 2012, 01:17
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If mv < pv< 0, is v > 0?

Given: $$mv<pv<0$$ --> two cases:

If $$v>0$$ then when dividing by $$v$$ we would have: $$m<p<0$$;
If $$v<0$$ then when dividing by $$v$$ we would have: $$m>p>0$$ (flip the sign when dividing by negative value).

(1) m < p --> we have the first case, so $$v>0$$. Sufficient.
(2) m < 0 --> we have the first case, so $$v>0$$. Sufficient.

Hope it's clear.
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Re: If mv < pv< 0, is v > 0? [#permalink]

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23 May 2012, 02:49
If mv < pv< 0, is v > 0?
from here, we know that:
mv < 0 -> if m is +, V must be -, if m is - then v is + (they must be opposite to be negative)
pv < 0 -> if p is +, V must be - ,if p is -, then v is + (they must be opposite to be negative)
mv < pv -> if v is +, then m < p,if v is - then m > p (divide both side by v,dont forget to flip inequality sign when v is -)

(1) m < p --> so v>0. Sufficient.
(2) m < 0 --> so v>0. Sufficient.
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Re: If mv < pv< 0, is v > 0? [#permalink]

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18 Oct 2014, 02:06
Bunuel wrote:
If mv < pv< 0, is v > 0?

Given: $$mv<pv<0$$ --> two cases:

If $$v>0$$ then when dividing by $$v$$ we would have: $$m<p<0$$;
If $$v<0$$ then when dividing by $$v$$ we would have: $$m>p>0$$ (flip the sign when dividing by negative value).

(1) m < p --> we have the first case, so $$v>0$$. Sufficient.
(2) m < 0 --> we have the first case, so $$v>0$$. Sufficient.

Hope it's clear.

I understood Bunuel's explanation for statement-1 but following values makes statement-1 insufficient. Please help me understand this:

(1) m<p

lets take v=1, m=-3, p=-2 it gives mv=-3, pv=-2 and hence does not violate mv<pv<0 as -3<-2<0, so v is +ve here
lets take v=-1, m=3, p=2 it gives mv=-3, pv=-2 and hence does not violate mv<pv<0 as -3<-2<0 but v is -ve here

Thanks
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Re: If mv < pv< 0, is v > 0? [#permalink]

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18 Oct 2014, 02:13
Expert's post
1
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HKD1710 wrote:
Bunuel wrote:
If mv < pv< 0, is v > 0?

Given: $$mv<pv<0$$ --> two cases:

If $$v>0$$ then when dividing by $$v$$ we would have: $$m<p<0$$;
If $$v<0$$ then when dividing by $$v$$ we would have: $$m>p>0$$ (flip the sign when dividing by negative value).

(1) m < p --> we have the first case, so $$v>0$$. Sufficient.
(2) m < 0 --> we have the first case, so $$v>0$$. Sufficient.

Hope it's clear.

I understood Bunuel's explanation for statement-1 but following values makes statement-1 insufficient. Please help me understand this:

(1) m<p

lets take v=1, m=-3, p=-2 it gives mv=-3, pv=-2 and hence does not violate mv<pv<0 as -3<-2<0, so v is +ve here
lets take v=-1, m=3, p=2 it gives mv=-3, pv=-2 and hence does not violate mv<pv<0 as -3<-2<0 but v is -ve here

Thanks

m = 3 and p = 2 violate the first statement, which says that m < p.
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Re: If mv < pv< 0, is v > 0? [#permalink]

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13 Mar 2016, 09:05
The golden rule to solve statement 1 is => inequality flips sign for mul/div with a negative variable else it retains its sign.
Hence D
Re: If mv < pv< 0, is v > 0?   [#permalink] 13 Mar 2016, 09:05
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