inequalities

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inequalities

by Carlo75 » Mon Jun 09, 2008 6:27 am
If mv < pv < 0, is v > 0 ?

(1) m < p

(2) m < 0

please explanations

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by zacharyz » Mon Jun 09, 2008 7:20 am
Question

For the two equations (mv and pv) to be less than zero, then one of the variables MUST be negative and one MUST be positive. So, either v is negative and both m and p are positive, or v is positive and both m and v are negative.

Statement 1 m < p

This doesn't give us anything useful on its own. they could both be positive or both be negative and we can't tell
Insufficient

Statement 2 m < 0
This says that m is negative. As I said under the question portion, if m is negative, then v must be positive (to make mv negative). You now know your answer. Sufficient

Thinking this through a little more, you know that pv is also negative (less than 0). v has to be positive, as it is the same variable as before. So that means that p has to be negative, to make pv negative. So, if the question had been asking is p > 0, you also have an answer for that.

The answer:
I. was insufficient - II was sufficient
(B)

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by ildude02 » Mon Jun 09, 2008 8:26 am
I think the answer is "D" since both are sufficent to determine that V >0.

When we consider (i), M and P can only take negative values to satisfy all the equations. So V shall be positive.We can quicky pick some numbers to make sure,

pick nubers such that m, p> 0 = (2, 3)
it saitsfies M < P(2< 3). But we also know that mv < pv <0. Therfore v< 0 to ensure that mv <0; so if v= -1, then substituing the vlaues for mv<pv, we get -2 <-3 ; but it doesnt satisfy mv < pv

cosnider m, p <0(-2, -3). It satisfies m< p(-3 <-2) ; it also satisfies that mv < pv <0> 0.

Please let me know if it's not true and may be I'm missing something.

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by Ian Stewart » Mon Jun 09, 2008 8:52 am
ildude- you're correct: Statement 1 is sufficient on its own. As you point out, by choosing numbers you do run the risk of 'missing something'. It's a good strategy to fall back on if you can't see how to do a problem abstractly or algebraically, but when we can do a problem abstractly, we can be sure we're getting the right answer. Here, using Statement 1 alone, we know:

m < p

Now, if we multiply both sides by v, we *must* have mv > pv if v is negative (if you multiply both sides of an inequality by a negative, you must reverse the inequality), and mv < pv if v is positive. Since we know, from the question, that mv < pv, we know that v must be positive.

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by zacharyz » Mon Jun 09, 2008 10:04 am
Great explanations. I fell for it. Thanks!

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by meet.anup » Mon Jun 09, 2008 11:38 pm
The ans is D.
from 1) v cannot be -ve.
if you substitute a -ve no for v and using m<p then mv<pv will not hold.
thus v has to be positive..

and 2) is suff also.

Thus, D

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by ildude02 » Sat Jun 14, 2008 2:41 pm
Ian Stewart wrote:ildude- you're correct: Statement 1 is sufficient on its own. As you point out, by choosing numbers you do run the risk of 'missing something'. It's a good strategy to fall back on if you can't see how to do a problem abstractly or algebraically, but when we can do a problem abstractly, we can be sure we're getting the right answer. Here, using Statement 1 alone, we know:

m < p

Now, if we multiply both sides by v, we *must* have mv > pv if v is negative (if you multiply both sides of an inequality by a negative, you must reverse the inequality), and mv < pv if v is positive. Since we know, from the question, that mv < pv, we know that v must be positive.
Great explanation! I jumped into numbers without thinking it from an "algebric" point'.