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Re: If mv < pv< 0, is v > 0? [#permalink]
21 May 2012, 00:17

2

This post received KUDOS

Expert's post

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.

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

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

Re: If mv < pv< 0, is v > 0? [#permalink]
18 Oct 2014, 01:13

Expert's post

1

This post was BOOKMARKED

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 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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