Solve for x, (x+m), and draw a Venn diagram for the given information.

Using the formula $|R \cup P| = |R| + |P| - |R \cap P|$:

$$19 = 13 + 19 - |R \cap P|$$ $$19 = 32 - |R \cap P|$$ $$|R \cap P| = 32 - 19$$ $$|R \cap P| = 13$$

Rulers only: $|R \setminus P| = |R| - |R \cap P| = 13 - 13 = 0$
Pencils only: $|P \setminus R| = |P| - |R \cap P| = 19 - 13 = 6$

i) Draw a Venn diagram to illustrate the information:
(A Venn diagram cannot be drawn in text, but the values for each region are provided below.)

• Pupils with rulers only: 0
• Pupils with pencils only: 6
• Pupils with both rulers and pencils: 13
• Pupils with neither: 6

ii) Find the number of pupils who have both rulers and pencils.
The number of pupils who have both rulers and pencils is $|R \cap P|$.

$$|R \cap P| = \boxed{13}$$

iii) Find the number of pupils who have pencil only.
The number of pupils who have pencil only is $|P \setminus R|$.

$$|P \setminus R| = \boxed{6}$$

iv) What percentage of the pupil have both ruler and pencil?
Percentage = $\frac{\text{Number with both}}{\text{Total pupils}} \times 100\%$

$$\text{Percentage} = \frac{13}{25} \times 100\%$$ $$\text{Percentage} = 0.52 \times 100\%$$ $$\text{Percentage} = \boxed{52\%}$$

Q2. (a)
Let the marked price be $M$.
The discount is 10% of the marked price, which is $0.10M$.
The selling price (amount paid) is GHS 810.
Selling Price = Marked Price - Discount

$$810 = M - 0.10M$$ $$810 = 0.90M$$ $$M = \frac{810}{0.90}$$ $$M = \boxed{\text{GHS } 900}$$

Q2. (b)

i) How many girls like Mathematics?
Total pupils in the school = 360.
Pupils who like Mathematics = $\frac{1}{10} \times 360 = 36$.
Half of those pupils who like Mathematics are girls.
Girls who like Mathematics = $\frac{1}{2} \times 36 = \boxed{18}$

ii) Find the truth set of $3x - 4 < 4x - 5$.

$$3x - 4 < 4x - 5$$

Subtract $3x$ from both sides:

$$-4 < 4x - 3x - 5$$ $$-4 < x - 5$$

Add 5 to both sides:

$$-4 + 5 < x$$ $$1 < x$$

So, $x > 1$.
The truth set is $\boxed{\{x \in \mathbb{R} \mid x > 1\}}$

iii) Make $p$ the subject of the relation $3t - pq = 2(P+L)$.

$$3t - pq = 2(P+L)$$

Expand the right side:

$$3t - pq = 2P + 2L$$

Isolate the term containing $p$:

$$-pq = 2P + 2L - 3t$$

Multiply both sides by -1:

$$pq = -(2P + 2L - 3t)$$ $$pq = 3t - 2P - 2L$$

Divide by $q$ (assuming $q \neq 0$):

$$p = \boxed{\frac{3t - 2P - 2L}{q}}$$