The average of 12 numbers is 18.5. The average of the first six numbers is 16.8 and that of the last seven numbers is 17.4. If the $$6^{th}$$ number is excluded, then what is the average (correct to one decimal place) of the remaining 11 numbers?

The average of 12 numbers is 18.5.

Sum of 12 numbers = $$12\times18.5$$ = 222

The average of the first six numbers is 16.8.

Sum of first six numbers = $$6\times16.8$$ = 100.8

The average of the last seven numbers is 17.4.

Sum of last seven numbers = $$7\times17.4$$ = 121.8

Sixth number = (Sum of first six numbers + Sum of last seven numbers) - Sum of 12 numbers

= (100.8+121.8)-222

= 222.6-222

= 0.6

If the $$6^{th}$$ number is excluded, then the average (correct to one decimal place) of the remaining 11 numbers = $$\frac{\left(222-0.6\right)}{11}$$

= $$\frac{221.4}{11}$$

= 20.1272727

= 20.1 (correct to one decimal place)