$$\frac{sum\ of\ 6\ numbers}{6}=8$$
$$sum\ of\ 6\ numbers\ =\ 8\ \cdot\ 6\ =\ 48$$
$$one\ additional\ positive\ integer\ is\ added$$
$$T\arg et\ question\ =>\ is\ the\ arithemetic\ mean\ of\ 7\ numbers\ greater\ than\ 10?$$
$$for\ the\ mean\ or\ average\ of\ 7\ numbers\ to\ be\ at\ least\ 10;$$
$$\frac{sum\ of\ 7\ numbers}{7}=10$$
$$sum\ of\ 7\ numbers\ =\ 70$$
$$for\ the\ sum\ of\ 7\ numbers\ to\ be\ =\ 70$$
$$new\ number\ must\ be\ =\ sum\ of\ 7\ numbers\ -\ sum\ of\ 6\ numbers$$
$$new\ number\ must\ be\ =\ 70-48=22$$
$$so\ if\ new\ number\ >\ 22\ then\ mean\ of\ 7\ numbers\ must\ be\ >\ 10$$
$$Statement\ 1\ =>\ the\ addditional\ integer\ is\ at\ least\ 23$$
$$this\ means\ new\ integer\ =\ 23\ and\ it\ is\ >\ 22\ so\ the\ mean\ of\ 7\ numbers\ >\ 10$$
$$statement\ 1\ is\ \ SUFFICIENT$$
$$Statement\ 2\ =>\ the\ additional\ integer\ is\ a\ multiple\ of\ 4$$
$$the\ view\ number\ can\ either\ be\ =\ 4,8,12,16,20,24\ etc\ which\ means\ it\ can\ be\ <\ 22\ or\ >\ 22$$
$$\sin ce\ this\ statement\ does\ not\ give\ definite\ value\ for\ new\ integer,$$
$$statement\ 2\ is\ NOT\ SUFFICIENT$$
$$Since\ statement\ 1\ alone\ is\ SUFFICIENT,\ \ answer\ =\ A$$
