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? 
How many teachers????? 
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Dan 
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Guys please help with the question 
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 $\neg P = 36$ $\neg X = 70$ $P \cap X = 61$ $(PX) \cup (XP) = 68$ $(PX) \cup (XP) = P + X  2P \cap X$ $NP = 36$ $P = N36$ $NX = 70$ $X = N70$ $(N36) + (N70)  2(61) = 68$ $2N = 296$ $N = 148$ 
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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