Answer: B
\(\therefore\) \(\cfrac{5+15+22+x+y+25+z}{7}=14\) Or, \(\cfrac{x+y+z+67}{7}=14\) Or, \(x+y+z+67=98\) Or, \(x+y+z=31\)
\(\therefore\) \(\cfrac{5+15+22+x+y+25+z}{7}=14\) Or, \(\cfrac{x+y+z+67}{7}=14\) Or, \(x+y+z+67=98\) Or, \(x+y+z=31\)