2016-I-12

$\begin{array}{rcl}
11 + a & = & 11 + b + 4 \\
a & = & b+ 4
\end{array}$

Note that $a > 11$ and $4\lt b < 10$. Therefore the possible values of $a$ and $b$ are $\left\{ \begin{array}{l} a=12 \\ b =8 \end{array} \right.$ or $\left\{ \begin{array}{l} a=13 \\ b =9 \end{array} \right.$.

1. Note that the ages of the four children are all different and the range of the ages of the children in the group remains unchanged. Then there are only two possibilities of the ages of the four children, they are $6$, $7$, $8$ and $9$ years old or $7$, $8$, $9$ and $10$ years old. For the greatest possible median of the ages of the children in the group, the suitable possibility is the four children are $7$, $8$, $9$ and $10$ years old. Therefore, the greatest possible median is $8$ years old.
2. For the least possible mean of the ages of the children in the group, the suitable possibility is the four children are $6$, $7$, $8$ and $9$ years old. Consider the two cases in part (a).

   For $a=12$ and $b=8$. Therefore, the mean of the ages of the children in the group

   $\begin{array}{cl}
   = & \dfrac{12\times 6 + 13 \times 7 + 12 \times 8 + 9 \times 9 + 4 \times 10}{12+ 13+ 12+ 9+ 4} \\
   = & 7.6
   \end{array}$

   For $a=13$ and $b=9$. Therefore, the mean of the ages of the children in the group

   $\begin{array}{cl}
   = & \dfrac{12\times 6 + 14 \times 7 + 12 \times 8 + 10 \times 9 + 4 \times 10}{12+ 14+ 12+ 10+ 4} \\
   = & 7.615\ 384\ 615
   \end{array}$

   Hence, the least possible mean of the ages of the children in the group is $7.6$.