SolUtion
D

Let the number of adults be $n$ and let the total of the ages of the children be $T$. The information in the question tells us that the mean age of the children is $20 \%$ or $\frac{1}{5}$ of the mean age of the whole group.

Therefore $\frac{1}{5} \times \frac{156+T}{3 n}=\frac{T}{2 n}$. Therefore $2(156+T)=3 \times 5 T$ and hence $312=13 C$. This has solution $T=24$.
20. There were twice as many children as adults sitting round a table. The age of each person at the table was a positive integer greater than 1. The sum of the ages of the adults was 156. The mean age of the children was $80 \%$ less than the mean age of the whole group.
What the sum of the ages of the children?
A 10
B 12
C 18
D 24
E 27