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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.

You can solve such questions easily by re-stating '< 0' as 'negative' and '> 0' as 'positive'.

mv < pv < 0 implies both 'pv' and 'mv' are negative and mv is more negative i.e. has greater absolute value as compared to pv. Since v will be equal in both, m will have a greater absolute value as compared to p.

When will mv and pv both be negative? In 2 cases: Case 1: When v is positive and m and p are both negative. Case 2: When v is negative and m and p are both positive.

So how will we know whether v is positive? If we know that at least one of m and p is negative, then v must be positive. If at least one of m and p is positive, then v must be negative.

Now that we understand the question and the implications of the given data, we go on to the statements.

Stmnt 1: m < p m has greater absolute value as compared to p but it is still smaller than p. This means m must be negative. If m is negative, p must be negative too which implies that v must be positive. Sufficient.

Stmnt 2: m < 0 Very straight forward. m and p both must be negative and v must be positive. Sufficient.

(1) means both m and p are negative, so in order \(mv\) and \(pv\) to be < 0, \(v\) must be greater than zero. (If it's -ve mv will > 0) (2) same is in (1) m<0 means \(m\) is -ve, and in order mv to be negative v must be greater than zero. Answer D
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Ask yourself: if m=3 and p=5 and v is negative, say -1, does mv < pv< 0 hold true?

Aha!! I get it now. So, when m=3, p=5 and v is -ve, mv (-3) becomes > pv (-5) making the given condition void.

So, Stmt 1 is sufficient. Great learning for the day. (This makes me wanna repeat to myself - When you pick numbers, quickly plug in to see if they are correct)

I also figured this just now:

mv < pv < 0 (mv-pv) <0 v(m-p)<0

If v is +ve, m<p (This is what the Statement 1 is saying too) If v is -ve, m>p

Statement 1) m < p I tried plugging numbers : m=-3, p=-2 to satisfy (a) consider different values of v : v is positive : v=5 , (-3)(5) < (-2)(5) < 0 = -15 < -10 < 0 ----- satisfy (a) v is negative : v=-5 , (-3)(-5) < (-2)(-5) < 0 = 15 < 10 < 0 ----- does not satisfy (a) Hence, v must be positive

Statement 2) m < 0 from (a) , we can see that mv < 0 hence to satisfy mv < 0 when m < 0 , we need a positive value of v [(-ve)*(+ve)=(-ve)] Therefore v must be positive

Could someone (@Bunuel) please check this alternative approach?

Rephrase stem to \(mv-pv<0\) --> \(v(m-p)<0\)

Stm 1: \(m<p\) --> \(m-p<0\), so \(v\) has to be positive for the above inequality to hold true. Sufficient.

Stm 2: Now this is where i screwed it up since i focused on my rephrased inequality and completely ignored the given one. Is there a way to draw the right conclusion from this inequality \(v(m-p)<0\) in combination with the constraint \(m<0\) of stm 2?

Otherwise i have to adjust my approach for those kind of questions since i tought rephrasing the question stem would in most cases help to evaluate both statements. Probably in this case it made things more complicated...