Simple ratio problem that I can't solve. One fifth of criminals are hard-core criminals. The hard-core criminals commit two-thirds of the criminal acts. What is the ratio of the number of criminal acts committed by the average hard-core criminal to the number commited by the average criminal who is not hard-core? P.S. I hope this sort of question isn't frowned upon since its very low level math (ACT test practice). If so let me know.

Since we have that $\frac{2}{3}$ of the acts are done by the hardcore the ones that are not is $\frac{3-2}{3}$

The average amount the hardcore criminals do is $\frac{5\times 2}{3}$(divide the number they do by percent of them there are). For regular it is $\frac{5\times 1}{4\times 3}$ so the ratio is $10:\frac{5}{4}$ Now multiply both sides by $\frac{4}{5}$ and you get $8:1$