The Venn diagram below shows the number of learners in a class who play netball, tennis, or volleyball. Given that there are 160 learners in the class.

*a) (i) The completed Venn diagram is as follows: Let N be Netball, T be Tennis, and V be Volleyball. The regions are filled as: Netball only: 40 Tennis only: 25 Volleyball only: 28 Netball and Tennis only (not Volleyball): 50 - x Netball and Volleyball only (not Tennis): 35 - x Tennis and Volleyball only (not Netball): 30 - x Netball, Tennis, and Volleyball (all three): x (ii) To find the value of x, we sum all the disjoint regions in the Venn diagram and equate it to the total number of learners, which is 160. 40 + 25 + 28 + (50 - x) + (35 - x) + (30 - x) + x = 160 Combine the constant terms: 40 + 25 + 28 + 50 + 35 + 30 = 208 Combine the terms with x: -x - x - x + x = -2x So the equation becomes: 208 - 2x = 160 Subtract 208 from both sides: -2x = 160 - 208 -2x = -48 Divide by -2: x = (-48)/(-2) x = 24 The value of x is 24. *b) Given the universal set = \x : 1 x < 10, x Z\. This means = \1, 2, 3, 4, 5, 6, 7, 8, 9\. A is a set of odd numbers. B is a set of factors of 10. (i) To list all elements of set A: Set A consists of odd numbers from the universal set . A = \1, 3, 5, 7, 9\ (ii) To find n(A' B): First, list the elements of set B. Factors of 10 are 1, 2, 5, 10. From , B is: B = \1, 2, 5\ Next, find the complement of A, denoted as A'. A' contains all elements in that are not in A. A' = \2, 4, 6, 8\ Now, find the intersection of A' and B, denoted as A' B. This set contains elements common to both A' and B. A' B = \2\ Finally, find n(A' B), which is the number of elements in the set A' B. n(A' B) = 1 The value of n(A' B) is 1. 3 done, 2 left today. You're making progress.