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(B) The correct option is $B$.Given that $R$ and $S$ are equivalence relations on set $A$.Since $R \subseteq A 	imes A$ and $S \subseteq A 	imes A$,

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$R \cap S$ is an equivalence relation on $A$

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(B) The correct option is $B$.Given that $R$ and $S$ are equivalence relations on set $A$.Since $R \subseteq A 	imes A$ and $S \subseteq A 	imes A$,it follows that $R \cap S \subseteq A 	imes A$,so $R \cap S$ is a relation on $A$.$1.$ Reflexivity: Let $a \in A$. Since $R$ and $S$ are reflexive,$(a, a) \in R$ and $(a, a) \in S$. Thus,$(a, a) \in R \cap S$ for all $a \in A$. So,$R \cap S$ is reflexive.$2.$ Symmetry: Let $(a, b) \in R \cap S$. Then $(a, b) \in R$ and $(a, b) \in S$. Since $R$ and $S$ are symmetric,$(b, a) \in R$ and $(b, a) \in S$. Thus,$(b, a) \in R \cap S$. So,$R \cap S$ is symmetric.$3.$ Transitivity: Let $(a, b) \in R \cap S$ and $(b, c) \in R \cap S$. Then $(a, b) \in R, (b, c) \in R$ and $(a, b) \in S, (b, c) \in S$. Since $R$ and $S$ are transitive,$(a, c) \in R$ and $(a, c) \in S$. Thus,$(a, c) \in R \cap S$. So,$R \cap S$ is transitive.Since $R \cap S$ is reflexive,symmetric,and transitive,it is an equivalence relation on $A$.

Let $R$ and $S$ be two equivalence relations on a set $A$. Then

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