A bin contains $10 \mathrm{~kg}$ of peanuts. $2 \mathrm{~kg}$ of peanuts are removed and $2 \mathrm{~kg}$ of raisins are added and thoroughly mixed in. Then $2 \mathrm{~kg}$ of this mixture are removed and $2 \mathrm{~kg}$ of raisins are added and thoroughly mixed in again. What is the ratio of the mass of peanuts to the mass of raisins in the final mixture?
(A) $3: 2$
(B) $4: 1$
(C) $5: 1$
(D) $7: 3$
(E) $16: 9$
When $2 \mathrm{~kg}$ of the $10 \mathrm{~kg}$ of peanuts are removed, there are $8 \mathrm{~kg}$ of peanuts remaining.

Since $2 \mathrm{~kg}$ of raisins are added, then there are $2 \mathrm{~kg}$ of raisins in the bin.

The peanuts and raisins are thoroughly mixed.

Since $2 \mathrm{~kg}$ of this mixture is removed and this is one-fifth of the total mass, then one-fifth of the mass of peanuts (or $\frac{8}{5} \mathrm{~kg}$ ) is removed and one-fifth of the mass of raisins (or $\frac{2}{5} \mathrm{~kg}$ ) is removed. This leaves $8-\frac{8}{5}=\frac{32}{5} \mathrm{~kg}$ of peanuts, and $2-\frac{2}{5}=\frac{8}{5} \mathrm{~kg}$ of raisins.

When $2 \mathrm{~kg}$ of raisins are added, the mass of raisins becomes $\frac{8}{5}+2=\frac{18}{5} \mathrm{~kg}$.

There are $\frac{32}{5} \mathrm{~kg}$ of peanuts and $\frac{18}{5} \mathrm{~kg}$ of raisins in the bin.

Therefore, the ratio of the masses is $\frac{32}{5}: \frac{18}{5}=32: 18=16: 9$.

ANSWER: (E)