3.3 In a class of 100 learners, 80 do Mathematics and 35 do Physics.
If x learners do Mathematics and Physics, and y learners do
neither Mathematics nor Physics, find the greatest possible value
of y.
Solution:
n(T)=100,n(M)=80,n(P)=35,n(M∩P)=x,n(M′∩P′)=y=n(M∪P)′⇒n(T)−n(M∪P)=y⇒100−n(M∪P)=y⇒n(M∪P)=100−yn(T)=100,n(M)=80,n(P)=35, \\n(M\cap P)=x, \\n(M'\cap P')=y=n(M\cup P)' \\\Rightarrow n(T)-n(M\cup P)=y \\\Rightarrow 100-n(M\cup P)=y \\\Rightarrow n(M\cup P)=100-yn(T)=100,n(M)=80,n(P)=35,n(M∩P)=x,n(M′∩P′)=y=n(M∪P)′⇒n(T)−n(M∪P)=y⇒100−n(M∪P)=y⇒n(M∪P)=100−y
Using n(M∪P)=n(M)+n(P)−n(M∩P)n(M \cup P)=n(M)+n(P)-n(M\cap P)n(M∪P)=n(M)+n(P)−n(M∩P)
⇒100−y=80+35−x⇒y=x−15\Rightarrow 100-y=80+35-x \\\Rightarrow y=x-15⇒100−y=80+35−x⇒y=x−15
Now, n(M∩P)≤n(P)n(M\cap P)\le n(P)n(M∩P)≤n(P)
⇒x≤35\Rightarrow x\le35⇒x≤35
Now for maximum value of yyy, xxx must be 35.
So, y=35−15=20y=35-15=20y=35−15=20
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