Chaitanya_1986 wrote:please help

Is m+z>0;
1)m-3z>0;
m>3z;
adding z on both sides we have;
m+z>4z;

consider z=1; m=5; m+z=6; which is greater than 4z and is also greater than zero;

consider z=-1; m=-2; m+z=-3; which is greater than 4z but is less than zero;

hence 1 alone is not sufficient to answer the question.

2)4z-m>0;
4z>m;
adding z on both sides we have; 5z>m+z;
z=1,m=3; m+z=4 which is less than 5z and is greater than zero;
z=-1 and m=-6 m+z=-7 which is less than 5z and is greater less than zero;
hence 2 alone is also not sufficient to answer the question..!!!

combining 1 and 2 we have;
m-3z>0; and 4z-m>0; also we know that when two positive quantities are added together their result will always be a positive;

adding these two equations we have; m-3z+4z-m>0;
z>0; as z is greater than zero hence we can safely conclude than sum of m+z will always be greater than zero hence answer should be C

Thanks Buddy....Gud explanation

Is m+z > 0?

(1) m-3z > 0
(2) 4z-m > 0

Statement 1: m > 3z.
m and z could both be positive, m and z could both be negative.
Insufficient.

Statement 2: m < 4z
m and z could both be positive, m and z could both be negative.
Insufficient.

Statements 1 and 2 combined:
Linking the inequalities, we get:
3z < m < 4z
3z < 4z.
0 < z.
Since z is positive, we know that m -- which is between 3z and 4z -- also is positive.
Thus, m+z > 0.
Sufficient.

The correct answer is C.