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If mv < pv< 0, is v > 0? [#permalink]
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20 May 2012, 13:14
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If mv < pv< 0, is v > 0? (1) m < p (2) m < 0
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Re: Is v positive? [#permalink]
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20 May 2012, 17:18
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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, 00:17
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Re: If mv < pv< 0, is v > 0? [#permalink]
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23 May 2012, 01:49
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, 01:06
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.
Answer: D.
Hope it's clear. I understood Bunuel's explanation for statement1 but following values makes statement1 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, 01:13
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.
Answer: D.
Hope it's clear. I understood Bunuel's explanation for statement1 but following values makes statement1 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, 08:05
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
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Re: If mv < pv< 0, is v > 0? [#permalink]
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29 Oct 2017, 12:34
Exactly similar question to this one from GMAT Prep :
If zy < xy < 0, is xz + x = z? (1) z < x (2) y > 0 Happy Prepping!




Re: If mv < pv< 0, is v > 0?
[#permalink]
29 Oct 2017, 12:34






