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Question-99606




Question Number 99606 by bemath last updated on 22/Jun/20
Commented by john santu last updated on 22/Jun/20
let M be the set of the students passing  in Mathematics , P be the set of students   passing in Physics & C be the set of   students passing in Chemistry  Now n(M∪P∪C) = 50 , n(M) = 37  n(P) = 24 , n(C)=43 , n(M∩P)≤19  n(M∩C)≤29 , n(P∩C)≤20  ⇒37+24+43−19−29−20+  n(M∩P∩C)≤50   ⇒n(M∩P∩C) ≤ 14 . Thus the largest  possible number that could have   passed all the three examination is 14 ■
$$\mathrm{let}\:\mathrm{M}\:\mathrm{be}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{the}\:\mathrm{students}\:\mathrm{passing} \\ $$$$\mathrm{in}\:\mathrm{Mathematics}\:,\:\mathrm{P}\:\mathrm{be}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{students}\: \\ $$$$\mathrm{passing}\:\mathrm{in}\:\mathrm{Physics}\:\&\:\mathrm{C}\:\mathrm{be}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\: \\ $$$$\mathrm{students}\:\mathrm{passing}\:\mathrm{in}\:\mathrm{Chemistry} \\ $$$$\mathrm{Now}\:\mathrm{n}\left(\mathrm{M}\cup\mathrm{P}\cup\mathrm{C}\right)\:=\:\mathrm{50}\:,\:\mathrm{n}\left(\mathrm{M}\right)\:=\:\mathrm{37} \\ $$$$\mathrm{n}\left(\mathrm{P}\right)\:=\:\mathrm{24}\:,\:\mathrm{n}\left(\mathrm{C}\right)=\mathrm{43}\:,\:\mathrm{n}\left(\mathrm{M}\cap\mathrm{P}\right)\leqslant\mathrm{19} \\ $$$$\mathrm{n}\left(\mathrm{M}\cap\mathrm{C}\right)\leqslant\mathrm{29}\:,\:\mathrm{n}\left(\mathrm{P}\cap\mathrm{C}\right)\leqslant\mathrm{20} \\ $$$$\Rightarrow\mathrm{37}+\mathrm{24}+\mathrm{43}−\mathrm{19}−\mathrm{29}−\mathrm{20}+ \\ $$$$\mathrm{n}\left(\mathrm{M}\cap\mathrm{P}\cap\mathrm{C}\right)\leqslant\mathrm{50}\: \\ $$$$\Rightarrow\mathrm{n}\left(\mathrm{M}\cap\mathrm{P}\cap\mathrm{C}\right)\:\leqslant\:\mathrm{14}\:.\:\mathrm{Thus}\:\mathrm{the}\:\mathrm{largest} \\ $$$$\mathrm{possible}\:\mathrm{number}\:\mathrm{that}\:\mathrm{could}\:\mathrm{have}\: \\ $$$$\mathrm{passed}\:\mathrm{all}\:\mathrm{the}\:\mathrm{three}\:\mathrm{examination}\:\mathrm{is}\:\mathrm{14}\:\blacksquare \\ $$

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