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Question Number 33663 by mondodotto@gmail.com last updated on 21/Apr/18

 Qn:(a)   given A and B are disjoint sets   shade in venn diagram   (i)B∩C (ii)A−(B∪C) (iii)(B∪C)−A   Qn:(b)   Given that ∣A∣=19 ∣B∣=23 ∣C∣=24   ∣A∪B∣=30 ∣(A∩B)−C∣=5 ∣(B∩C)∣=10   ∣(A∪B)^′ ∣=20 and ∣C−(A∪B)∣=10   then find   (i)∣(A∩C)∣ (ii)∣(B−′C)∣′   (iii)∣𝛍∣ (iv)∣(B′∪C)∩A′∣

$$\:\boldsymbol{\mathrm{Qn}}:\left(\boldsymbol{\mathrm{a}}\right) \\ $$$$\:\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{B}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{disjoint}}\:\boldsymbol{\mathrm{sets}} \\ $$$$\:\boldsymbol{\mathrm{shade}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{venn}}\:\boldsymbol{\mathrm{diagram}} \\ $$$$\:\left(\boldsymbol{\mathrm{i}}\right)\boldsymbol{\mathrm{B}}\cap\boldsymbol{\mathrm{C}}\:\left(\boldsymbol{\mathrm{ii}}\right)\boldsymbol{\mathrm{A}}−\left(\boldsymbol{\mathrm{B}}\cup\boldsymbol{\mathrm{C}}\right)\:\left(\boldsymbol{\mathrm{iii}}\right)\left(\boldsymbol{\mathrm{B}}\cup\boldsymbol{\mathrm{C}}\right)−\boldsymbol{\mathrm{A}} \\ $$$$\:\boldsymbol{\mathrm{Qn}}:\left(\boldsymbol{\mathrm{b}}\right) \\ $$$$\:\boldsymbol{\mathrm{G}}\mathrm{i}\boldsymbol{\mathrm{ven}}\:\boldsymbol{\mathrm{that}}\:\mid\boldsymbol{\mathrm{A}}\mid=\mathrm{19}\:\mid\boldsymbol{\mathrm{B}}\mid=\mathrm{23}\:\mid\boldsymbol{\mathrm{C}}\mid=\mathrm{24} \\ $$$$\:\mid\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}\mid=\mathrm{30}\:\mid\left(\boldsymbol{\mathrm{A}}\cap\boldsymbol{\mathrm{B}}\right)−\boldsymbol{\mathrm{C}}\mid=\mathrm{5}\:\mid\left(\boldsymbol{\mathrm{B}}\cap\boldsymbol{\mathrm{C}}\right)\mid=\mathrm{10} \\ $$$$\:\mid\left(\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}\right)^{'} \mid=\mathrm{20}\:\boldsymbol{\mathrm{and}}\:\mid\boldsymbol{\mathrm{C}}−\left(\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}\right)\mid=\mathrm{10} \\ $$$$\:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{find}} \\ $$$$\:\left(\boldsymbol{\mathrm{i}}\right)\mid\left(\boldsymbol{\mathrm{A}}\cap\boldsymbol{\mathrm{C}}\right)\mid\:\left(\boldsymbol{\mathrm{ii}}\right)\mid\left(\boldsymbol{\mathrm{B}}−'\boldsymbol{\mathrm{C}}\right)\mid' \\ $$$$\:\left(\boldsymbol{\mathrm{iii}}\right)\mid\boldsymbol{\mu}\mid\:\left(\mathrm{iv}\right)\mid\left(\boldsymbol{\mathrm{B}}'\cup\boldsymbol{\mathrm{C}}\right)\cap\boldsymbol{\mathrm{A}}'\mid \\ $$

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