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In a group which occurs by people who play guitar and battery, The value of who plays guitar is $2$ times of who plays battery. The value of who plays only an instrument is $12$ and who plays the both instrument is $3$.

  • How do you find how many person does only play the battery?

I'm so surprised for this question. What would the true equation be?

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    $\begingroup$ Can you please make some attempt? Something like Jorge did for your previous question? $\endgroup$ – Bram28 Oct 31 '17 at 18:15
  • $\begingroup$ @Bram28 I was going to show my attempt but didn't know where to start. $\endgroup$ – user440404 Oct 31 '17 at 18:40
  • $\begingroup$ OK, but with fleablood's and my hints, can you get the right equations now? $\endgroup$ – Bram28 Oct 31 '17 at 18:43
  • $\begingroup$ @Bram28 I'm trying to get it at the moment. $\endgroup$ – user440404 Oct 31 '17 at 18:45
  • $\begingroup$ OK, let me know if you need more help. $\endgroup$ – Bram28 Oct 31 '17 at 18:46
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Break it down into distinct sets.

There are 1) People who play Battery. That is set $B$.

There are 2) People who play Guitar. That is set $G$.

There are 3) People who play Both. That is set $BG = B \cap G$. We know $|BG| =3$.

There are 4) People who only play Battery. That is the set $BO = B \setminus (B \cap G)$ and $BO \cup (B\cap G) = B$ and $BO \cap (B\cap G) = \emptyset$. So $|BO| = |B| - |BG| = |B| - 3$. and $|BO| + |BG| = |B|= |BO| + 3$.

There are 5) People who only play Guitar That is the set $GO = G \setminus (B \cap G)$ and $GO \cup (B\cap G) = G$ and $GO \cap (B\cap G) = \emptyset$. So $|GO| = |G| - |BG| = |G| - 3$. and $|BGO| + |BG| = |G|= |GO| + 3$.

And there are 6) People who only play one instrument. That is $O = (B \cup G) \setminus (B \cap G) = BO \cup GO$. We know $BO \cap GO = \emptyset$ and $BO \cup GO = O$ and so we know $|O| = |BO| + |GO|$ and we are told $|O| = 12$.

We are told:

$|BG| = 3$, $|O| = 12$, and $|G| = 2|B|$.

Then its just a matter of solving.

enter image description here[![enter image description here]1

$|G| = 2|B|$

$|GO| + |B\cap G| = 2(|BO| + |B\cap G|)$

$|GO| + 3 = 2(|BO| + 3)$ and $|BO| + |GO| = 12$.

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  • $\begingroup$ Can you show it a bit more? $\endgroup$ – user440404 Oct 31 '17 at 18:57
  • $\begingroup$ So $G = 2B; GO + BO = 12; BG = 3$. $\endgroup$ – fleablood Oct 31 '17 at 19:02
  • $\begingroup$ I know that $BG = 3$ How did you get others? If you show them on your answer, I would be too happy. $\endgroup$ – user440404 Oct 31 '17 at 19:02
  • $\begingroup$ Can you please show it? How did you find others? I'm getting that $G = GO +3$ and $B = BO +3$. $\endgroup$ – user440404 Oct 31 '17 at 19:14
  • $\begingroup$ I really need your help atm. $\endgroup$ – user440404 Oct 31 '17 at 19:45
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HINT

Here's another Venn Diagram:

enter image description here

We know 3 people play both guitar and drums (no one in the English-speaking part of the world plays battery :) )

Say $X$ people play guitar but not battery and $Y$ people play drums but not guitar. Can you now translate the given pieces of information into equations?

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  • $\begingroup$ no one in the English-speaking part of the world plays battery :).... Well, my cousin Tony who lives in the garage.... we don't talk about him much.... $\endgroup$ – fleablood Oct 31 '17 at 18:30
  • $\begingroup$ @fleablood LOL. OK, so one can play battery ...but not for long :) $\endgroup$ – Bram28 Oct 31 '17 at 18:34
  • $\begingroup$ I got $6+2x$ for $G$ $\endgroup$ – user440404 Oct 31 '17 at 19:50