Show that $A \cap B=A \cap C$ need not imply $B = C$
If $A \cap B = B$, then
If $A = \{x : x$ is a multiple of $4\}$ and $B = \{x : x$ is a multiple of $6\}$ then $A \cap B$ consists of all multiples of
In a school, there are three types of games to be played. Some of the students play two types of games, but none play all the three games. Which Venn diagrams can justify the above statement ?
If $A =$ [$x:x$ is a multiple of $3$] and $B =$ [$x:x$ is a multiple of $5$], then $A -B$ is ($\bar A$ means complement of $A$)