$$Total\ number\ of\ people\ in\ a\ group\ \ge10$$
$$Regular\ adult\ fee\ is\ now\ =\ 5500\ -\ \left(10\%\cdot5500\right)$$
$$=¥4950$$
$$Regular\ child\ fee\ =\ 4800\ -\left(10\%\cdot4800\right)=4320$$

Target question => How many children are on the group?

Statement 1 => The total of the admission fee paid for the adults in the group was ¥29700
If 1 adult fee = 4950
then x adult fee = 29700
x = 29700/4950 = 6
This does not give us any specific information relating to the children so the target question cannot be answered. Statement 1 is NOT SUFFICIENT

Statement 2 => The total of the admission fee paid for the children in the group was ¥4860 more than the total of the admission fee paid for the adults in the group
Let the number of adults = a and number of children = c
Total adult fee = 4950a and total children fee = 4320c
4320c = 4950a + 4860 but the exact value of a or c is unknown, hence statement 2 is NOT SUFFICIENT

Combining both statements together =>
Let the number of adults = a and number of children = c
From statement 1 => a = 6
From statement 2 => 4320 = 4950a + 4860
Substituting the value of a from statement 1 in the equation in statement 2
4320c = 4950(6) + 4860
4320c/4320 = 34560/4320
c = 8
There are 8 children in the group. Both statements together ARE SUFFICIENT

Answer = C