# 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.

• Any math questions are acceptable as long as they are well thought out. – ruler501 Mar 23 '14 at 2:06

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 $8:1$

• The resource I'm using claims the answer is 8:1 with this explanation: The hard-core criminals commit twice as much crime (two thirds is twice as big as one third), with only a quarter as many people (one fifth are hard core, four fifths are not hard core). So that's eight times as much crime per person. – Atlas Mar 23 '14 at 2:08
• They simplify to the same thing. I'll add in another step to show that it is the same. – ruler501 Mar 23 '14 at 2:09
• Alright thanks that clears it up. That's awful I didn't recognize it though. – Atlas Mar 23 '14 at 2:14
• It wasn't obvious to me at first either. – ruler501 Mar 23 '14 at 2:15

Hardcore= (2/3)/(1/5)

TheRest= (1/3)/(4/5)

Hardcore/TheRest = 8

• Very nice straight to the point explanation. – Atlas Mar 23 '14 at 2:14