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

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



Statement 1 : Since m<p
(m-p)<0
We also know that mv<pv ie (m-p)v<0
Since (m-p)<0 therefore v>0
SUFFICIENT

Statement 2: Given m<0
Since mv<0
therefore v > 0
SUFFICIENT

Hence (D)
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Re: If mv < pv < 0, is v > 0? [#permalink]
I am sorry but I somehow still dont understand. I chose B, which I know is incorrect.

Well, let me tell you why I do not understand the Stmt. 1 is sufficient.

Given: mv<pv<0. It is given that MV and PV ARE -ve. This means, When V is -ve, M,P are +ve and vice versa.

I tabulated as below:

m v p mv pv
+ - + - -
- + - - -

Now, statement 1 says m < p. It does not say if they are negative or positive.

So, it is possible that:

3 < 5 (m=3 and p=5) and this means V is -ve

OR

-3 < -1 (m=-3 and p=-1) and this means V is +ve

Different answers so stmt 1 should be insufficient. What I am missing?

Thank you!
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Re: If mv < pv < 0, is v > 0? [#permalink]
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flower07 wrote:
I am sorry but I somehow still dont understand. I chose B, which I know is incorrect.

Well, let me tell you why I do not understand the Stmt. 1 is sufficient.

Given: mv<pv<0. It is given that MV and PV ARE -ve. This means, When V is -ve, M,P are +ve and vice versa.

I tabulated as below:

m v p mv pv
+ - + - -
- + - - -

Now, statement 1 says m < p. It does not say if they are negative or positive.

So, it is possible that:

3 < 5 (m=3 and p=5) and this means V is -ve


OR

-3 < -1 (m=-3 and p=-1) and this means V is +ve

Different answers so stmt 1 should be insufficient. What I am missing?

Thank you!


Ask yourself: if m=3 and p=5 and v is negative, say -1, does mv < pv< 0 hold true?
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Re: If mv < pv < 0, is v > 0? [#permalink]
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Bunuel wrote:
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

So, the answer is D.

Thank you!!
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Re: If mv < pv< 0, is v > 0? [#permalink]
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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
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Re: If mv < pv< 0, is v > 0? [#permalink]
Expert Reply
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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If mv < pv < 0, is v > 0? [#permalink]
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Stiv wrote:
If mv < pv < 0, is v > 0?

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


(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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Re: If mv < pv < 0, is v > 0? [#permalink]
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I believe it is D.

Case 1) m < p
mv < pv < 0
mv - pv < pv - pv < -pv . Subtract pv
v(m-p) < 0 < -pv

Since m < p OR (m - p) < 0 Therefore, v must be positive. SUFF

Case 2) m < 0
Since mv < 0 (given), v must be positive. SUFF

Hence D.
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Re: If mv < pv< 0, is v > 0? [#permalink]
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enigma123 wrote:
If mv < pv< 0, is v > 0?

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



Main Topic


Inequalities

Rule:
Multiplying/dividing inequalities by
+ variable: Don't flip sign
- variable: Flip sign

Divide by v
If v = +, then m < p < 0
If v = -, then m > p > 0

1) Implies v = +. Sufficient.
2) Implies v = +. Sufficient.

ANSWER: D
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Re: If mv < pv< 0, is v > 0? [#permalink]
Expert Reply
enigma123 wrote:
If mv < pv< 0, is v > 0?

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


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. Then we recheck the question. We should simplify conditions if necessary.

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

Since the original condition is equivalent to \(v(m-p) < 0\), the question is equivalent to \(m - p < 0\) or \(m < p\).
This is same as condition 1).
Thus condition 1) is sufficient.

Condition 2)
When we consider both condition 2), \(m < 0\) and the original condition, \(mv < 0\), we have \(v > 0\).
Thus condition 2) is also sufficient.

Therefore, D is the answer.
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If mv < pv < 0, is v > 0? [#permalink]
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.


Damn it's solvable in less than 30 seconds! Never thought it was that easy! Thanks Bunuel!!!
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If mv < pv < 0, is v > 0? [#permalink]
If \(mv < pv < 0,\) is \(v > 0\)?

\(mv - pv < 0\)
\(v(m-p) < 0\)
\((m - p)\) and v must have opposite signs. For v to be positive, \((m - p)\) must be negative. We can rephrase the question as: Is \(m < p\)?

(1) \(m < p\)

Directly answers our question. SUFFICIENT.

(2) \(m < 0\)

Lets look back at the original question prompt: \(mv < pv < 0\)

If m is negative, v MUST be negative for the prompt to be true. Otherwise, mv would be greater than 0. SUFFICIENT.

Answer is D.
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Re: If mv < pv < 0, is v > 0? [#permalink]
Did not really like the solutions I read, so here is mine.
We know mv < pv < 0, so this means we know mv < 0 (aka negative) AND pv < 0 (aka negative).

(1) m < p

Let us say m = -5 and p = -1
Then -5v < -1v < 0 is only true if v is positive
Say v = 1, then -5 < -1 < 0 TRUE
Say v = -1, then 5 < 1 < 0 FALSE

Well, what if m = 1 and p = 5, it is still true m < p
Then 1v < 5v < 0
If v is negative, say v = -1 then
-1 < -5 < 0 FALSE so that is not possible
If v is positive, save v = 1 then
1 < 5 < 0 also FALSE

Let us say m = -1 and p = 3, again m < p
Then -1v < 3v < 0
if v is negative, say v = -1 then
1< -3 < 0 FALSE, not possible
if v is positive, say v = 1 then
-1 < 3 < 0 FALSE not positive

So I just proved to you that if m < p, v MUST be positive

(2) m < 0

Remember when I said mv < 0 (aka negative)?

if mv MUST be negative, and m is negative, then that means v MUST be positive since (-)(+) is (-)

So both are sufficient. This is more theory-based than using formulas. It is helpful to know both ways!!
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Re: If mv < pv < 0, is v > 0? [#permalink]
My 2 cents on this:

It seems better to not use numbers, since algebraic equations might suffice. I solved it with this logic:

Before heading to Statements, we are given mv< pv < 0. I'm quickly able to deduce that mv < 0 and pv < 0. Let's keep this in mind.

S1: m<p --> written now as m-p< 0

from main statement, we know mv<pv<0 --> mv - pv < 0 further written as v(m-p) < 0

Now lets look at v (m-p) < 0

either v < 0 or m-p < 0; from S1 we can confirm m-p< 0 and so v > 0 has to be true. Sufficient.

S2: m<0. Ding ding ding! Our initial deduction was mv<0 and if m< 0 v has to be positive. Sufficient.

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


This is such an elegant and simple solution. Your brain really is something else...
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