In a class of $80$ children, $35$% children can play only cricket, $45$% children can play only table-tennis and the remaining children can play both the games. In all, how many children can play cricket?
If x belongs to set of integers, A is the solution set of $2(x-1)< 3x-1$ and B is the solution set of $4x-3\leq 8+x$, find A$\cap$B.
If $A = {1, 3, 5, 7, 8, 6}$, $B = {2, 4, 6, 8, 9}$ .Find $A\cap B$
If $A = \left {2, 3, 4, 8, 10\right }, B = \left (3, 4, 5, 10, 12\right }, C = \left {4, 5, 6, 12, 14\right }$, then $(A\cap B)\cup (A\cap C)$ is equal to
If U = {1, 2, 3, 4, 5, 6}; A = {3, 5}; B = {2, 3, 4} C = {4, 5}, find then A $\cap$ (B $\cup$ C).
Which is the simplified representation of
$\left( {{A^/} \cap \,{B^/} \cap \,C} \right) \cup \left( {B\, \cap \,C} \right) \cup \left( {A \cap \,C} \right)$
where A,B,C are subsets of X
If $aN=\left{ ax:x\epsilon N \right}$, then the set $3N\cap 7N$ is
If $n(U)= 60, n(A)= 35, n(B)= 24$ and $n(A \cup B)' = 10$ ,then $n(A \cap B)$ is
A is a set containing $n$ elements. $A$ subset $P$ of $A$ is chosen. the set $A$ is reconstructed by replacing the elements of $P.A$ subset $Q$ of $A$ is again chosen. the number of ways of choosing $P$ and $Q$ so that $P \cap Q$
Let $A=\left{ a,b,c,d \right} ,B=\left{ b,c,d,e \right}$. Then $n\left[ \left( A\times B \right) \cap \left( B\times A \right) \right]$ is equal to