jmc

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 $d=\boxed{6}$, and $d$ is the number that appeared twice. Yay! :D