mv<pv<0
v>0??
a) m<p
mv<pv
(p-m)v>0
(p-m)>0
Thus v>0
Sufficent
b)m<0
mv<0
thus v>0
Sufficient
IMO D

Anurag@Gurome wrote:

divya23 wrote:if mv<pv<0 is v>0
m<p
m<0

Statement 1: m < p --> (m - p) < 0 --> (m - p) is negative
From question stem, mv < pv --> (mv - pv) < 0 --> v(m - p) < 0
As product of v and (m - p) is negative and (m - p) is also negative, v must be positive.

Sufficient.

Statement 2: m < 0 --> m is negative
From question stem, mv < 0
As product of v and m is negative and m is also negative, v must be positive.

Sufficient.

The correct answer is D.

Hi Anurag ,

As far as what i understand , for inequalities , if we are not aware of the sign of the variable , multiplication or division is not possible .

Can you please quantify on how you arrived at mv < pv from stem 1 . Are we not supposed to reverse the sign if 'v' has a different sign ?

Hi Anurag

Please could you also explain how we got to this stage

Statement 1: m < p --> (m - p) < 0 --> (m - p) is negative

How can you -p away from both sides when we do not know the sign for p?
Thank you

abasirov wrote:Hi Anurag

Please could you also explain how we got to this stage

Statement 1: m < p --> (m - p) < 0 --> (m - p) is negative

How can you -p away from both sides when we do not know the sign for p?
Thank you

Hi abasirov ,

I just understood the problem .

There's one rule that ul have to keep in mind for inequalities.
Rule - The sign of the inequality changes only during division or multiplication.

For eg : x + 1 < 2.

Here you are allowed to subtract -1 from both sides without a change in sign . So the equation becomes x + 1 - 1 < 2 - 1
x < 1 .

b . -x + 1 < 2

Following the same step as in ex 1 , we arive at -x < 1 . However we want the value of x , hence divide by -1 on both sides .This is where a sign change is mandated . The final equation becomes

x > -1 .

hope it helps.