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November 15th, 2018, 12:01 PM  #1 
Newbie Joined: Nov 2018 From: Indianapolis Posts: 18 Thanks: 0  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 12:14 PM. 
November 15th, 2018, 12:55 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670 
How many teachers?????

November 15th, 2018, 02:56 PM  #3  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,079 Thanks: 845 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
November 15th, 2018, 09:51 PM  #4 
Newbie Joined: Nov 2018 From: Indianapolis Posts: 18 Thanks: 0  Sorry I meant boys not teachers

November 15th, 2018, 10:28 PM  #5 
Newbie Joined: Nov 2018 From: Indianapolis Posts: 18 Thanks: 0 
Guys please help with the question

November 15th, 2018, 11:23 PM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,372 Thanks: 1275 
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 $\neg P = 36$ $\neg X = 70$ $P \cap X = 61$ $(PX) \cup (XP) = 68$ $(PX) \cup (XP) = P + X  2P \cap X$ $NP = 36$ $P = N36$ $NX = 70$ $X = N70$ $(N36) + (N70)  2(61) = 68$ $2N = 296$ $N = 148$ 
November 16th, 2018, 03:15 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
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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