There are $200$ individuals with a skin disorder, $120$ had been exposed to the chemical $C _{1}, 50$ to chemical $C _{2},$ and $30$ to both the chemicals $C _{1}$ and $C _{2} .$ Find the number of individuals exposed to
Chemical $C_{2}$ but not chemical $C_{1}$
Let $U$ denote the universal set consisting of individuals suffering from the skin disorder, $A$ denote the set of individuals exposed to the chemical $C_{1}$ and $B$ denote the set of individuals exposed to the chemical $C_{2}$
Here $\quad n( U )=200, n( A )=120, n( B )=50$ and $n( A \cap B )=30$
From the Fig we have
$B=(B-A) \cup(A \cap B)$
and so, $\quad n( B )=n( B - A )+n( A \cap B )$
( Since $B - A$ and $A \cap B$ are disjoint .)
or $n(B - A) = n(B) - n(A \cap B)$
$ = 50 - 30 = 20$
Thus, the number of individuals exposed to chemical $C_{2}$ and not to chemical $C_{1}$ is $20 .$
There are $200$ individuals with a skin disorder, $120$ had been exposed to the chemical $C _{1}, 50$ to chemical $C _{2},$ and $30$ to both the chemicals $C _{1}$ and $C _{2} .$ Find the number of individuals exposed to
Chemical $C_{1}$ or chemical $C_{2}$
In a survey of $220$ students of a higher secondary school, it was found that at least $125$ and at most $130$ students studied Mathematics; at least $85$ and at most $95$ studied Physics; at least $75$ and at most $90$ studied Chemistry; $30$ studied both Physics and Chemistry; $50$ studied both Chemistry and Mathematics; $40$ studied both Mathematics and Physics and $10$ studied none of these subjects. Let $\mathrm{m}$ and $\mathrm{n}$ respectively be the least and the most number of students who studied all the three subjects. Then $\mathrm{m}+\mathrm{n}$ is equal to .............................
A survey shows that $73 \%$ of the persons working in an office like coffee, whereas $65 \%$ like tea. If $x$ denotes the percentage of them, who like both coffee and tea, then $x$ cannot be
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