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
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inequalities
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Cans!!
Statement 1: m < p --> (m - p) < 0 --> (m - p) is negativedivya23 wrote:if mv<pv<0 is v>0
m<p
m<0
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.
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Hi Anurag ,Anurag@Gurome wrote:Statement 1: m < p --> (m - p) < 0 --> (m - p) is negativedivya23 wrote:if mv<pv<0 is v>0
m<p
m<0
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.
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
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 ,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
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.
















