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

[Luci]

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

(1) m<p
(2) m<0

GMAT PREP says the correct answer is D. But I thought it was B, I dont understand how 1 makes it sufficient.

(1) m<p and mv<pv<0
if m and p are negative values (ex. m=-5, p=-3) then v must be positive whatever is its value
if m and p are positive values (ex. m=3, p=5) then v must be negative if we have mv<pv<0
if m is negative and p is positive (ex. m=-2, p=3) then it is inconsistent with mv<pv<0

So I dont understand where I making the mistake

(2) m<0 and mv<pv<0

Here if m<0 then if we want mv<pv<0, v must be positive. SUFFICIENT

Thanks

[GMAT 2007]

(1) if m<p, mv<pv<0 will only hold if V is +ve and M and P are -ve. For other two cases that you mentioned mv>pv so they are not valid.

Hence, for mv<pv<0 V to hold, V has to +ve. So (1) is sufficient.

GMAT 2007

[StaceyKoprince]

Luci, you were on the right track with your process, but you didn't actually work out the numbers, so you didn't notice that some of them were inconsistent with the given condition (mv<pv<0). I love trying numbers as a technique, but make sure you follow through just a bit more on the calculations.

If you had, you would have seen:
ex. m=3, p=5, and let's make v =-2
mv = -6
pv = -10
-6<-10<0
Which is not true, so that combination is invalid - I can't use it to test the statement. The only way to make it work would be to make p less than m, but statement one gives the condition m<p, so I can't do it.

[harshasks]

Easier way to do this is

since mv<pv<0

Just take mv<pv => v(m-p)<0 , hence either v<0 or m-p<0 both cannot be true together.

Since in A you are given m<p => m-p<0 which makes v +ve always and hence A is sufficient .

You know B already.

[RonPurewal]

Easier way to do this is

since mv<pv<0

Just take mv<pv => v(m-p)<0 , hence either v<0 or m-p<0 both cannot be true together.

Since in A you are given m<p => m-p<0 which makes v +ve always and hence A is sufficient .

this is correct, although i would take out "easier" and replace it with "different". (people from different educational backgrounds will have widely differing ideas of what is "easier".)

You know B already.

not sure what this means.
if it means "this has already been derived above", then yes. if it means "this is given in the statement itself", then, not quite -- you still have to do some mathematics in order to derive it.

[ven2]

I worked it out in the following way

plug in numbers

consider possibility 1)

m =3 p=2 v=-5

2) m =-3 p=-2 v=5

both the above possibilities satisfy the equation mv<pv<0

Case 1) m<p so 1 is not possible so v as to be greater than 0

case 2) m<0 so yes only m =-3 p=-2 v=5 holds true



D

[jnelson0612]

Great! Yes, plugging numbers is such a useful way to handle problems such as this one.

[rachelhong2012]

You can also solve this problem without plugging in #'s. I did this problem with flow chart approach learned from advanced quant strategy book.

Basically, if I can rephrase the question starting with a condition:
that if v is positive, then I can divide v across the board directly.

This condition thus leads to two possiblities, one in which:
v =pos
so it becomes m<p<0

And one in which
v= neg
in which case m>p>0 (pretending that the scenario is opposite, then I had just flipped the sign by dividing by a neg number, hence I'm compensating it by flipping the sign later.

Therefore, we arrived at two different conclusions, based on whether v is positve or negative.

(1) m<p

this statement tells us that it's the first scenario of m<p<0 in which case v=pos
suff

(2) m<0
this statement also tells us that it's the first scenario of m<p<0 in which case v=pos
suff

hence D

[jnelson0612]

Rachel, great work! Thanks!

[shubham_sagijain]

I used the same approach as is used by Rachel...if a guy knows the rules of inequality, it is easier to work out this problem
But, i agree with Stacey that plugging numbers is always the safest bet :)

Thanks,
Shubh

[RonPurewal]

the safest bet is to have as many approaches as possible to the problem, without prejudice toward (or against) any particular method.

[PudGe]

Ron,Stacy..every one! i cant thank you all enough.. that was so much learning from one single post!

Cheers to you'all :D

[tim]

glad to hear it!