If $X$ and $Y$ are two sets such that $X \cup Y$ has $50$ elements, $X$ has $28$ elements and $Y$ has $32$ elements, how many elements does $X$ $\cap$ $Y$ have?
If $X$ and $Y$ are two sets such that $X \cup Y$ has $50$ elements, $X$ has $28$ elements and $Y$ has $32$ elements, how many elements does $X$ $\cap$ $Y$ have?
Given that
$n( X \cup Y )=50, n( X )=28, n( Y )=32$
$n( X \cap Y )=?$
By using the formula
$n( X \cup Y )=n( X )+n( Y )-n( X \cap Y ),$
we find that
$ n( X \cap Y ) =n( X )+n( Y )-n( X \cup Y ) $
$=28+32-50=10 $
Alternatively, suppose $n( X \cap Y )=k,$ then
$n( X - Y )=28-k, n( Y - X )=32-k$ (by Venn diagram in Fig )
This gives $50=n( X \cup Y )=n( X - Y )+n( X \cap Y )+n( Y - X )$
$=(28-k)+k+(32-k)$
Hence $k=10$
Similar Questions
If $A=\{3,5,7,9,11\}, B=\{7,9,11,13\}, C=\{11,13,15\}$ and $D=\{15,17\} ;$ find
$A \cap C \cap D$
If $A=\{3,5,7,9,11\}, B=\{7,9,11,13\}, C=\{11,13,15\}$ and $D=\{15,17\} ;$ find
$A \cap C \cap D$
Find sets $A, B$ and $C$ such that $A \cap B, B \cap C$ and $A \cap C$ are non-empty sets and $A \cap B \cap C=\varnothing$
Find sets $A, B$ and $C$ such that $A \cap B, B \cap C$ and $A \cap C$ are non-empty sets and $A \cap B \cap C=\varnothing$
The shaded region in given figure is-
The shaded region in given figure is-
If $A = \{1, 2, 3, 4, 5\}, B = \{2, 4, 6\}, C = \{3, 4, 6\},$ then $(A \cup B) \cap C$ is
If $A = \{1, 2, 3, 4, 5\}, B = \{2, 4, 6\}, C = \{3, 4, 6\},$ then $(A \cup B) \cap C$ is
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
$C-A$
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
$C-A$