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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.

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  • $\begingroup$ Any math questions are acceptable as long as they are well thought out. $\endgroup$ – ruler501 Mar 23 '14 at 2:06
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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 $8:1$

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    $\begingroup$ 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. $\endgroup$ – Atlas Mar 23 '14 at 2:08
  • $\begingroup$ They simplify to the same thing. I'll add in another step to show that it is the same. $\endgroup$ – ruler501 Mar 23 '14 at 2:09
  • $\begingroup$ Alright thanks that clears it up. That's awful I didn't recognize it though. $\endgroup$ – Atlas Mar 23 '14 at 2:14
  • $\begingroup$ It wasn't obvious to me at first either. $\endgroup$ – ruler501 Mar 23 '14 at 2:15
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Hardcore= (2/3)/(1/5)

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

Hardcore/TheRest = 8

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

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