The correct option is (B): 30Given that: arithmetic mean of \(x \space and \space y= 40\)\(⇒\)\(\frac{x+y}{2}=40\) ⇒ \(x+y=80\)given that \(z=10\)so A.M of \(x,y,z =\frac{x+y+z}{3}\)\(⇒\)\(\frac{80+10}{3}\)\(= 30\)