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Hint: - The following can be done by set theory as well as by creating charts but using set theory is quite a simpler way to solve.
(i) College people having heights more than five feet. These boys should belong to
both $S$ and $W$ as they have to satisfy both the properties.
$\therefore S \cap W$
(ii) People who are not teenagers and have their heights less than five feet. Teenagers belong
To $T$. Those who are not teenagers belong to $T'$. Similarly $W'$ and hence
$T' \cap W' = {\left\{ {T \cup W} \right\}^\prime }$
(iii) All people who are neither males nor teenagers nor college students
$M' \cap T' \cap S' = {\left\{ {M \cup T \cup S} \right\}^\prime }$
Note: - The signs used above are signs of set theory, where $ \cup $ represents union of two sets, $ \cap
$ represents the intersection of two sets and a bar like sign on the top of any letter represents the
conjugate of the set. (Like this:$A'$ )
(i) College people having heights more than five feet. These boys should belong to
both $S$ and $W$ as they have to satisfy both the properties.
$\therefore S \cap W$
(ii) People who are not teenagers and have their heights less than five feet. Teenagers belong
To $T$. Those who are not teenagers belong to $T'$. Similarly $W'$ and hence
$T' \cap W' = {\left\{ {T \cup W} \right\}^\prime }$
(iii) All people who are neither males nor teenagers nor college students
$M' \cap T' \cap S' = {\left\{ {M \cup T \cup S} \right\}^\prime }$
Note: - The signs used above are signs of set theory, where $ \cup $ represents union of two sets, $ \cap
$ represents the intersection of two sets and a bar like sign on the top of any letter represents the
conjugate of the set. (Like this:$A'$ )
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