The average of twelve numbers is 46. The average of the first four numbers is 43 and that of the last five numbers is 49.4.The $$5^{th}$$ and the $$6^{th}$$ numbers are respectively 4 and 6 more than the $$7^{th}$$ number. What is the average of the $$5^{th}$$ and the $$7^{th}$$ numbers?

Given that,

The average of the twelve number is =46,

Let the numbers are $$n_1, n_2, n_3....n_{12}$$

So, $$\dfrac{n_1+ n_2+ n_3....n_{12}}{12}=46$$

$$\Rightarrow n_1+ n_2+ n_3....n_{12}=46\times 12=552$$---------------------(i)

As per the given condition,

Average of first 4 number $$\dfrac{n_1+n_2+n_3+n_4}{4}=43$$

$$\Rightarrow n_1+n_2+n_3+n_4=43\times 4 =172$$----------------------(ii)

The average of last 5th number $$\dfrac{n_8+n_9+n_{10}+n_{11}+n_{12}}{5}=49.4$$

$$\Rightarrow n_8+n_9+n_{10}+n_{11}+n_{12}=49.4\times 5 =247.0$$-----------------(iii)

Let the 7th number is x, so 5$$\th$$ and 6$$\th$$ number$$x+4$$ and $$x+6$$ respectively.

Now from the equation (i) (ii) and (iii),

$$\Rightarrow n_1+ n_2+ n_3+n_4+(x+4)+(x+6)+x+n_8+n_9+n_{10}+n_{11}+n_{12}=552$$

$$\Rightarrow 172+(x+4)+(x+6)+x+247=552$$

$$\Rightarrow 172+3x+10+247=552$$

$$\Rightarrow 3x=552-429=123$$

$$\Rightarrow x=41$$

So, 5$$th$$ number is $$=41+4=45$$

Hence, the average of 5$$th$$ and 7$$th$$ number $$\dfrac{41+45}{2}=43$$

Hence, the required average number is 43.