The average of $10$ numbers is $41$. The average of first $3$ numbers is $39.4$ and the average of last $4$ numbers is $43.5$. $4^{\text{th}}$ number is $2$ more than $5^{\text{th}}$ number and $0.2$ more than $6^{\text{th}}$ number. What is the average of the $4^{\text{th}}$ and $6^{\text{th}}$ numbers?

Total sum of $10$ numbers $=10\times 41=410$.Sum of first $3$ numbers $=3\times 39.4=118.2$.Sum of last $4$ numbers $=4\times 43.5=174$.Sum of the remaining numbers ($4^{\text{th}}$, $5^{\text{th}}$, and $6^{\text{th}}$ numbers):
$$\text{Sum}=410-(118.2+174)=410-292.2=117.8$$
Let the $4^{\text{th}}$ number be $x$.According to the problem:$4^{\text{th}}$ number $=x$$5^{\text{th}}$ number $=x-2$$6^{\text{th}}$ number $=x-0.2$ Sum of these three numbers:
$$x+(x-2)+(x-0.2)=117.8$$
$$3x-2.2=117.8$$
$$3x=120$$
$$x=40$$
So,$4^{\text{th}}$ number $=40$$6^{\text{th}}$ number $=40-0.2=39.8$ Average of $4^{\text{th}}$ and $6^{\text{th}}$ numbers:
$$\text{Average}=\frac{40+39.8}{2}=\frac{79.8}{2}=39.9$$