$\frac{2^1+2^0+2^{-1}}{2^{-2}+2^{-3}+2^{-4}}$ equals 
$\text{(A)} \ 6 \qquad  \text{(B)} \ 8 \qquad  \text{(C)} \ \frac{31}{2} \qquad  \text{(D)} \ 24 \qquad  \text{(E)} \ 512$
We can factorise to give $\frac{2^1 + 2^0 + 2^{-1}}{2^{-3}(2^1 + 2^0 + 2^{-1})} = \frac{1}{2^{-3}} = 8$, which is $\boxed{\text{B}}$.