The problem describes the number of grade 12 learners who collected different types of textbooks: Biology, Mathematics, and History. We are given: - Total learners: 50 - Biology textbooks: 33 - Mathematics textbooks: 24 - History textbooks: 23 - Biology and Mathematics: 13 - Mathematics and History: 16 - Biology and History: 15 - All three types: 10 The problem asks us to: (i) Illustrate this information on a Venn diagram. (ii) Find: (a) How many learners collected none of the books. (b) How many learners collected only one type of textbook. (c) How many learners collected only two types of textbooks.

1. Problem Description

The problem describes the number of grade 12 learners who collected different types of textbooks: Biology, Mathematics, and History. We are given:

- Total learners: 50

- Biology textbooks: 33

- Mathematics textbooks: 24

- History textbooks: 23

- Biology and Mathematics: 13

- Mathematics and History: 16

- Biology and History: 15

- All three types: 10

The problem asks us to:

(i) Illustrate this information on a Venn diagram.

(ii) Find:

(a) How many learners collected none of the books.

(b) How many learners collected only one type of textbook.

(c) How many learners collected only two types of textbooks.

2. Solution Steps

Let B, M, and H represent the sets of learners who collected Biology, Mathematics, and History textbooks, respectively. We are given the following:

$|B| = 33$

$|M| = 24$

$|H| = 23$

$|B \cap M| = 13$

$|M \cap H| = 16$

$|B \cap H| = 15$

$|B \cap M \cap H| = 10$

(i) Venn Diagram:

We can fill in the Venn diagram as follows:

- $|B \cap M \cap H| = 10$

- $|B \cap M| - |B \cap M \cap H| = 13 - 10 = 3$

- $|M \cap H| - |B \cap M \cap H| = 16 - 10 = 6$

- $|B \cap H| - |B \cap M \cap H| = 15 - 10 = 5$

- $|B| - (|B \cap M| + |B \cap H| - |B \cap M \cap H|) = 33 - (13 + 15 - 10) = 33 - 18 = 15$

- $|M| - (|B \cap M| + |M \cap H| - |B \cap M \cap H|) = 24 - (13 + 16 - 10) = 24 - 19 = 5$

- $|H| - (|B \cap H| + |M \cap H| - |B \cap M \cap H|) = 23 - (15 + 16 - 10) = 23 - 21 = 2$

(ii)

(a) Number of learners who collected none of the books:

Total number of learners who collected at least one book:

$|B \cup M \cup H| = |B| + |M| + |H| - |B \cap M| - |B \cap H| - |M \cap H| + |B \cap M \cap H|$

$|B \cup M \cup H| = 33 + 24 + 23 - 13 - 15 - 16 + 10 = 80 - 44 + 10 = 36 + 10 = 46$

Number of learners who collected none: $50 - 46 = 4$

(b) Number of learners who collected only one type of textbook:

Only Biology: 15

Only Mathematics: 5

Only History: 2

Total: $15 + 5 + 2 = 22$

(c) Number of learners who collected only two types of textbooks:

Only Biology and Mathematics: 3

Only Mathematics and History: 6

Only Biology and History: 5

Total: $3 + 6 + 5 = 14$

3. Final Answer

(i) Venn Diagram: (Values are calculated in the solution steps)

B only = 15, M only = 5, H only = 2

B and M only = 3, M and H only = 6, B and H only = 5

B and M and H = 10

(ii)

(a) Number of learners who collected none of the books: 4

(b) Number of learners who collected only one type of textbook: 22

(c) Number of learners who collected only two types of textbooks: 14