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November 15th, 2018, 01:01 PM   #1
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Set Theory Question

Hard set theory question please help ASAP

Out of a group of boys, 36 don’t have PS4, 70 don’t have an XBOX, whilst 61 have both a PS4 and an XBOX. If 68 boys have one or other machine, but not both, then how many teachers are in the group?

Last edited by Jeff Shreeves; November 15th, 2018 at 01:14 PM.
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November 15th, 2018, 01:55 PM   #2
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How many teachers?????
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November 15th, 2018, 03:56 PM   #3
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Quote:
Originally Posted by Jeff Shreeves View Post
Hard set theory question please help ASAP

Out of a group of boys, 36 don’t have PS4, 70 don’t have an XBOX, whilst 61 have both a PS4 and an XBOX. If 68 boys have one or other machine, but not both, then how many teachers are in the group?
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Originally Posted by mathman View Post
How many teachers?????
Well, one teacher took away 36 PS4s because the students weren't paying attention and the second teacher took away 70 XBOXs because the students stuck gum under their desks. So two teachers.

-Dan
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November 15th, 2018, 10:51 PM   #4
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Quote:
Originally Posted by Jeff Shreeves View Post
Hard set theory question please help ASAP

Out of a group of boys, 36 don’t have PS4, 70 don’t have an XBOX, whilst 61 have both a PS4 and an XBOX. If 68 boys have one or other machine, but not both, then how many teachers are in the group?
Sorry I meant boys not teachers
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November 15th, 2018, 11:28 PM   #5
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Guys please help with the question
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November 16th, 2018, 12:23 AM   #6
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Let $N$ be the total number of boys

let $P$ be the set of boys that have a PS4

let $X$ be the set of boys that have an Xbox

we are given that

$|\neg P| = 36$
$|\neg X| = 70$
$|P \cap X| = 61$
$|(P-X) \cup (X-P)| = 68$

$|(P-X) \cup (X-P)| = |P| + |X| - 2|P \cap X|$

$N-|P| = 36$
$|P| = N-36$
$N-|X| = 70$
$|X| = N-70$

$(N-36) + (N-70) - 2(61) = 68$

$2N = 296$

$N = 148$
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November 16th, 2018, 04:15 AM   #7
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As 36 + 70 = 68 + 2 × number of boys with neither machine, 19 boys have neither machine.

Hence total number of boys = 61 + 68 + 19 = 148.
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