$$\left(x+y\right)z>y+z$$
Open bracket
$$xz+yz>y+2$$
$$xz+yz-z>y$$ $$\left(x+y-1\right)2>y$$ $$Is\left(x+y\right)z\ >y+z\ \ OR\ is\ \left(x+y-1\right)z>y?$$
Statement 1
This means the least value of x is 1 since it is a positive integer ( greater than 0). So when x=1. then yz is equal to y which directly means (x+y)z is equal to (y+z) but if z=2
$$2y>\ y$$
This is indirectly means (x+y)z > y+z
$$Statement\ \ 1\ is\ NOT\ SUFFICIENT.$$
Statement 2 2>x
This means the least value of x is 1 and least value of 2 is 2 so, when x=1 and z = 2
$$\left(x+y-1\right)z>y\ =\ \left(1+y-1\right)2>y$$ $$=zy>y$$ $$Hence,\ \left(x+y\right)z>y+z\ Statement\ 2\ is\ SUFFICIENT.$$ $$Answer\ is\ Option\ B$$
