Question

Let $A=\{1,2,3,4\},$ $B=\{4,5,6,7\},$ $C=\{3,6,9,12\}$ and $D=\{4,8,12,16,20\},$ then $n\left[\right(A\times B)\cap (C\times D\left)\right]$ =

• A. $2$
• B. $4$
• C. $1$
• D. $8$

Answer 1

The correct option is C $1$

$A\times B=\left\{\right(1,4),(1,5),(1,6),(1,7),(2,4),(2,5),(2,6),(2,7),(3,4),(3,5),(3,6),(3,7),(4,4),(4,5),(4,6),(4,7\left)\right\}$
And, $C\times D=\left\{\right(3,4),(3,8),(3,12),(3,16),(3,20),(6,4),(6,8),(6,12),(6,16),(6,20),(9,4),(9,8),(9,12),(9,16),(9,20),(12,4),(12,8),(12,12),(12,16),(12,20\left)\right\}$
Now, $(A\times B)\cap (C\times D)=\{3,4\}$
$\therefore n\left[\right(A\times B)\cap (C\times D\left)\right]=1$