Convert the division into multiplication by inverting the second fraction.

Cancel 5, 7, $a^2$, $b$, $c$, $d^2$:

The final answer is $\boxed{\frac{\text{9cd}}{\text{ab}}}$.

1 (b) Solve the equation $2x^2 - x - 2 = 0$, giving your answers correct to 2 decimal places.

Here, $a = 2$, $b = -1$, $c = -2$.

For $x_1$:

For $x_2$:

The final answers are $\boxed{\text{x = 1.28}}$ and $\boxed{\text{x = -0.78}}$.

2 (a) Two boxes A and B contain balls of the same size. Box A contains 2 green balls and 3 blue balls while Box B contains 4 green balls and 2 blue balls. If one ball is selected at random from each box, what is the probability that both balls selected are

• Box A: Total balls = $2 \text{ (green)} + 3 \text{ (blue)} = 5$ $P(\text{Green from A}) = \frac{2}{5}$ $P(\text{Blue from A}) = \frac{3}{5}$
• Box B: Total balls = $4 \text{ (green)} + 2 \text{ (blue)} = 6$ $P(\text{Green from B}) = \frac{4}{6} = \frac{2}{3}$ $P(\text{Blue from B}) = \frac{2}{6} = \frac{1}{3}$

(i) blue

The final answer is $\boxed{\text{}\frac{1}{5}\text{}}$ or $\boxed{\text{0.2}}$.

(ii) of different colours?

Scenario 1: Green from Box A AND Blue from Box B.
Scenario 2: Blue from Box A AND Green from Box B.

The final answer is $\boxed{\text{}\frac{8}{15}\text{}}$.

2 (b) On a particular day at a school with 50 Grade 12 learners, 33 collected Biology textbooks, 24 collected Mathematics textbooks and 23 collected History textbooks from the school library. 13 learners collected both Biology and Mathematics textbooks, 16 collected both Mathematics and History textbooks, 15 collected both Biology and History textbooks and 10 collected all the three types of textbooks.

Let B = Biology, M = Mathematics, H = History.
Total learners = 50.
$|B| = 33$
$|M| = 24$
$|H| = 23$
$|B \cap M| = 13$
$|M \cap H| = 16$
$|B \cap H| = 15$
$|B \cap M \cap H| = 10$

(i) Illustrate this information in a Venn diagram.

To construct the Venn diagram, we first fill in the innermost intersection and work outwards.

• All three (B $\cap$ M $\cap$ H): 10
• Only B and M: $|B \cap M| - |B \cap M \cap H| = 13 - 10 = 3$
• Only M and H: $|M \cap H| - |B \cap M \cap H| = 16 - 10 = 6$
• Only B and H: $|B \cap H| - |B \cap M \cap H| = 15 - 10 = 5$

Now, calculate those who collected only one type of textbook:

• Only B: $|B| - (\text{Only B and M}) - (\text{Only B and H}) - (|B \cap M \cap H|) = 33 - 3 - 5 - 10 = 15$
• Only M: $|M| - (\text{Only B and M}) - (\text{Only M and H}) - (|B \cap M \cap H|) = 24 - 3 - 6 - 10 = 5$
• Only H: $|H| - (\text{Only M and H}) - (\text{Only B and H}) - (|B \cap M \cap H|) = 23 - 6 - 5 - 10 = 2$

Venn Diagram Regions:

• 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) How many learners collected

(a) none of the textbooks,

This is the sum of all regions within the Venn diagram.

The final answer is $\boxed{\text{4}}$.

(b) one type of textbook only,

The final answer is $\boxed{\text{22}}$.

(c) two types of textbooks only?

The final answer is $\boxed{\text{14}}$.