There is a list of seven numbers. The average of the first four numbers is $5$, and the average of the last four numbers is $8$. If the average of all seven numbers is $6\frac{4}{7}$, then the number common to both sets of four numbers is
$\text{(A)}\ 5\frac{3}{7}\qquad\text{(B)}\ 6\qquad\text{(C)}\ 6\frac{4}{7}\qquad\text{(D)}\ 7\qquad\text{(E)}\ 7\frac{3}{7}$
Remember that if a list of $n$ numbers has an average of $k$, then the sum $S$ of all the numbers on the list is $S = nk$.
So if the average of the first $4$ numbers is $5$, then the first four numbers total $4 \cdot 5 = 20$.
If the average of the last $4$ numbers is $8$, then the last four numbers total $4 \cdot 8 = 32$.
If the average of all $7$ numbers is $6\frac{4}{7}$, then the total of all seven numbers is $7 \cdot 6\frac{4}{7} = 7\cdot 6 + 4 =  46$.
If the first four numbers are $20$, and the last four numbers are $32$, then all "eight" numbers are $20 + 32 = 52$.  But that's counting one number twice.  Since the sum of all seven numbers is $46$, then the number that was counted twice is $52 - 46 = 6$, and the answer is $\boxed{(B) 6}$
Algebraically, if $a + b + c + d = 20$, and $d + e + f + g = 32$, you can add both equations to get $a + b + c + 2d + e + f + g = 52$.  You know that $a + b + c + d + e + f + g = 46$, so you can subtract that from the last equation to get $d = \boxed{(B) 6}$, and $d$ is the number that appeared twice.
Yay!  :D