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# Let U= {4, 8, 12, 16, 20, 24, 28}, A= { 8, 16, 24} and B= { 4, 16, 20, 28}. Find ${\text{(A}} \cup {\text{B)'}}$ and ${\text{(A}} \cap {\text{B)'}}$.

Last updated date: 24th Jul 2024
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Hint: In order to solve this problem we need to know that $\cup \,{\text{and }} \cap$ are the signs of union and intersection. Union means collection of all the elements of all the sets provided and intersection between two sets means the element common in those sets will be considered and if there is a set P then P’ means all the elements except that of P. Knowing this information you will get the right answer to this question.

We have the set U= {4, 8, 12, 16, 20, 24, 28}, A= { 8, 16, 24} and B= { 4, 16, 20, 28}.
We have to find ${\text{(A}} \cup {\text{B)'}}$ and ${\text{(A}} \cap {\text{B)'}}$
We know that ${\text{(A}} \cup {\text{B)}}$ means all the elements in A and B
Therefore, ${\text{(A}} \cup {\text{B)}}$={4, 8, 16, 20, 24, 28}
And ${\text{(A}} \cup {\text{B)'}}$ means all the elements of U except that present in ${\text{(A}} \cup {\text{B)}}$.
Therefore, ${\text{(A}} \cup {\text{B)'}}$= {12}

We know that ${\text{(A}} \cap {\text{B)}}$ means all the elements common in A and B.
Therefore, ${\text{(A}} \cap {\text{B)}}$={16}
And ${\text{(A}} \cap {\text{B)'}}$ means all the elements of U except that present in ${\text{(A}} \cap {\text{B)}}$.
Therefore, ${\text{(A}} \cap {\text{B)'}}$= {4, 8, 12, 20, 24, 28}.
Hence, ${\text{(A}} \cup {\text{B)'}}$= {12} and ${\text{(A}} \cap {\text{B)'}}$= {4, 8, 12, 20, 24, 28}.

Note: Whenever you face such types of problems you need to know that $\cup \,{\text{and }} \cap$ are the signs of union and intersection. Union means collection of all the elements of all the sets provided and intersection between two sets means the element common in those sets will be considered and if there is a set P then P’ means all the elements except that of P. Proceeding with the help of this will solve your problem.