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. $$4^{th}$$ number is 2 more than the $$5^{th}$$ number and 0.2 more than the $$6^{th}$$ number. What is the average of the $$4^{th}$$ and $$6^{th}$$ numbers

Average of 10 numbers = 41

=> Sum of 10 numbers = $$41\times10=410$$

Similarly, sum of first 3 numbers = $$39.4\times3=118.2$$

Sum of last 4 numbers = $$43.5\times4=174$$

Let 4th, 5th and 6th numbers be $$x,y,z$$ respectively.

=> $$x+y+z=410-118.2-174=117.8$$ ----------------(i)

Also, $$x=y+2=z+0.2=k$$ (say) ------------(ii)

Solving above equations, we get : $$k+(k-2)+(k-0.2)=117.8$$

=> $$3k=117.8+2.2=120$$

=> $$k=\frac{120}{3}=40$$

Thus, $$x=40,y=38,z=39.8$$

$$\therefore$$ Average of the $$4^{th}$$ and $$6^{th}$$ numbers = $$\frac{40+39.8}{2}=39.9$$

=> Ans - (D)