Question

If $A=\{3,5,7,9,11\},B=\{7,9,11,13\}$
$C=\{11,13,15\}$ and $D=\{15,17\}$;
$\left(i\right)A\cap B$
$\left(ii\right)B\cap C$
$\left(iii\right)A\cap C\cap D$
$\left(iv\right)A\cap C$
$\left(v\right)B\cap D$
$\left(vi\right)A\cap (B\cup C)$
$\left(vii\right)A\cap D$
$\left(viii\right)A\cap (B\cup D)$
$\left(ix\right)(A\cap B)\cap (B\cup C)$
$\left(x\right)(A\cup D)\cap (B\cup C)$

Answer 1

Given: $A=\{3,5,7,9,11\},B=\{7,9,11,13\}$
$C=\{11,13,15\}$ and $D=\{15,17\}$

$\therefore A\cap B=\{3,5,7,9,11\}\cap \{7,9,11,13\}$
Hence, $A\cap B=\{7,9,11\}$

$\therefore B\cap C=\{7,9,11,13\}\cap \{11,13,15\}$
Hence, $B\cap C=\{11,13\}$

$\because A\cap C\cap D=(A\cap C)\cap D$
$\because A\cap C=\left\{11\right\}$
Now, $A\cap C\cap D=\left\{11\right\}\cap \{15,17\}$
Hence, $A\cap C\cap D=\varphi$

$\therefore A\cap C=\{3,5,7,9,11\}\cap \{11,13,15\}$
Hence, $A\cap C=\left\{11\right\}$

$\therefore B\cap D=\{7,9,11,13\}\cap \{15,17\}$
Hence, $B\cap D=\varphi$

$\because A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$
$\because A\cap B=\{7,9,11\}\text{and}A\cap C=\left\{11\right\}$
Now, $A\cap (B\cup C)=\{7,9,11\}\cup \left\{11\right\}$
Hence, $A\cap (B\cup C)=\{7,9,11\}$

$\therefore A\cap D=\{3,5,7,9,11\}\cap \{15,17\}$
Hence, $A\cap D=\varphi$

$\because A\cap (B\cup D)=(A\cap B)\cup (A\cap D)$
$\because A\cap B=\{7,9,11\}\text{and}A\cap D=\varphi$
$\therefore A\cap (B\cup D)=\{7,9,11\}\cup \varphi$
Hence, $A\cap (B\cup D)=\{7,9,11\}$

$\because A\cap B=\{7,9,11\}\text{and}B\cup C=\{7,9,11,13,15\}$
$\therefore (A\cap B)\cap (B\cup C)=\{7,9,11\}\cap \{7,9,11,13,15\}$
Hence, $(A\cap B)\cap (B\cup C)=\{7,9,11\}$

$\because A\cup D=\{3,5,7,9,11,15,17\}\text{and}B\cup C=\{7,9,11,13,15\}$
$\therefore (A\cap D)\cap (B\cup C)=\{3,5,7,9,11,15,17\}\cap \{7,9,11,13,15\}$
Hence, $(A\cup D)\cap (B\cup C)=\{7,9,11,15\}$