If $X$ and $Y$ are two sets such that $X$ has $40$ elements, $X \cup Y$ has $60$ elements and $X$ $\cap\, Y$ has $10$ elements, how many elements does $Y$ have?
It is given that:
$n(X)=40, n(X \cup Y)=60, n(X \cap Y)=10$
We know that:
$n(X \cup Y)=n(X)+n(Y)-n(X \cap Y)$
$\therefore 60=40+n(Y)-10$
$\therefore n(Y)=60-(40-10)=30$
Thus, the set $Y$ has $30$ elements.
If $A=\{3,6,9,12,15,18,21\}, B=\{4,8,12,16,20\},$ $C=\{2,4,6,8,10,12,14,16\}, D=\{5,10,15,20\} ;$ find
$A-C$
If $A = \{ x:x$ is a natural number $\} ,B = \{ x:x$ is an even natural number $\} $ $C = \{ x:x$ is an odd natural number $\} $ and $D = \{ x:x$ is a prime number $\} ,$ find
$B \cap D$
Find the union of each of the following pairs of sets :
$A = \{ x:x$ is a natural number and multiple of $3\} $
$B = \{ x:x$ is a natural number less than $6\} $
If $aN = \{ ax:x \in N\} $ and $bN \cap cN = dN$, where $b$, $c \in N$ are relatively prime, then
If $A=\{3,6,9,12,15,18,21\}, B=\{4,8,12,16,20\},$ $C=\{2,4,6,8,10,12,14,16\}, D=\{5,10,15,20\} ;$ find
$A-B$