Venn diagrams and complement questions with answers

Questions

1. Question 1Read a set from a Venn diagram

   The Venn diagram shows the elements of $\xi$, $A$ and $B$. List the elements of $A'$.

   Show the worked solution

   1. Step 1: Which regions make up the set? Yes or No for each$A'$: not in A.Regions: B only, outside both circles.
   2. Step 2: Collect the elements in those regions$A' = \{1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18\}$
2. Question 2Read a set from a Venn diagram

   The Venn diagram shows the elements of $\xi$, $A$ and $B$. List the elements of $(A \cap B)'$.

   Show the worked solution

   1. Step 1: Which regions make up the set? Yes or No for each$(A \cap B)'$: not in both.Regions: A only, B only, outside both circles.
   2. Step 2: Collect the elements in those regions$(A \cap B)' = \{5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20\}$
3. Question 3Which region is shaded?

   Which diagram shows the set $A \cup B$ shaded?

   Show the worked solution

   1. Step 1: Which regions make up the set? Yes or No for each$A \cup B$: in A or B (or both).Regions: A only, the overlap, B only.
   2. Step 2: Find the diagram with exactly those regions shadedDiagram A.
4. Question 4Which region is shaded?

   Which diagram shows the set $A' \cap B$ shaded?

   Show the worked solution

   1. Step 1: Which regions make up the set? Yes or No for each$A' \cap B$: in B but not in A.Regions: B only.
   2. Step 2: Find the diagram with exactly those regions shadedDiagram A.
5. Question 5Name the shaded set

   Which set is shaded in the Venn diagram?

   Show the worked solution

   1. Step 1: Which regions are shaded? Yes or No for eachShaded: the overlap.
   2. Step 2: Find the set made of exactly those regionsThat is $A \cap B$: in both A and B.
6. Question 6Name the shaded set

   Which set is shaded in the Venn diagram?

   Show the worked solution

   1. Step 1: Which regions are shaded? Yes or No for eachShaded: A only, outside both circles.
   2. Step 2: Find the set made of exactly those regionsThat is $B'$: not in B.