A ratio is a comparison of two or more quantities of the same kind, expressed as a fraction or with a colon. It shows how much of one quantity there is compared to another.

$$
\text{Total parts} = 3 (\text{boys}) + 2 (\text{girls}) = 5 \text{ parts}
$$

$$
\text{Value per part} = \frac{\text{Total students}}{\text{Total parts}} = \frac{30}{5} = 6 \text{ students/part}
$$

$$
\text{Number of boys} = \text{Boys' ratio part} \times \text{Value per part} = 3 \times 6 = 18
$$
There are $\boxed{\text{18 boys}}$.

$$
\text{Hourly rate} = \frac{\text{Total earnings}}{\text{Total hours}} = \frac{\text{GHC}600}{8 \text{ hours}} = \text{GHC}75/\text{hour}
$$
The hourly rate of pay is $\boxed{\text{GHC}75 \text{ per hour}}$.

$$
\text{Total parts} = 5 (\text{male}) + 7 (\text{female}) = 12 \text{ parts}
$$

$$
\text{Value per part} = \frac{\text{Total students}}{\text{Total parts}} = \frac{1200}{12} = 100 \text{ students/part}
$$

$$
\text{Number of female students} = \text{Female ratio part} \times \text{Value per part} = 7 \times 100 = 700
$$
There are $\boxed{\text{700 female students}}$ in the school.

$$
\text{Actual distance} = \text{Map distance} \times \text{Scale factor}
$$
$$
\text{Actual distance} = 8 \text{ cm} \times \frac{5 \text{ km}}{1 \text{ cm}} = 40 \text{ km}
$$
The actual distance between the towns is $\boxed{\text{40 km}}$.

$$
\text{Total worker-days} = 6 \text{ workers} \times 15 \text{ days} = 90 \text{ worker-days}
$$

$$
\text{Days needed} = \frac{\text{Total worker-days}}{\text{Number of workers}} = \frac{90 \text{ worker-days}}{9 \text{ workers}} = 10 \text{ days}
$$
It would take $\boxed{\text{10 days}}$ for 9 workers to complete the same job.

$$
\text{Total parts} = 2 + 3 + 5 = 10 \text{ parts}
$$

$$
\text{Value per part} = \frac{\text{Total profit}}{\text{Total parts}} = \frac{\text{GHC}5000}{10} = \text{GHC}500/\text{part}
$$

Friend 1: $2 \times \text{GHC}500 = \text{GHC}1000$  
Friend 2: $3 \times \text{GHC}500 = \text{GHC}1500$  
Friend 3: $5 \times \text{GHC}500 = \text{GHC}2500$  
The friends receive $\boxed{\text{GHC}1000}$, $\boxed{\text{GHC}1500}$, and $\boxed{\text{GHC}2500}$ respectively.

$$
\text{Fuel efficiency} = \frac{120 \text{ km}}{8 \text{ litres}} = 15 \text{ km/litre}
$$

$$
\text{Litres needed} = \frac{300 \text{ km}}{15 \text{ km/litre}} = 20 \text{ litres}
$$
$\boxed{\text{20 litres}}$ of fuel are needed to travel 300 km.

$$
\text{Total parts} = 3 (\text{cows}) + 4 (\text{goats}) + 5 (\text{sheep}) = 12 \text{ parts}
$$

$$
\text{Value per part} = \frac{\text{Total animals}}{\text{Total parts}} = \frac{60}{12} = 5 \text{ animals/part}
$$

Cows: $3 \times 5 = 15$ cows  
Goats: $4 \times 5 = 20$ goats  
Sheep: $5 \times 5 = 25$ sheep  
Reasoning: The total number of animals (60) is divided by the sum of the ratio parts (12) to find the value of one part (5 animals). This value is then multiplied by each individual ratio part to determine the count for each animal type, ensuring the proportions are maintained.  
The farmer has $\boxed{\text{15 cows}}$, $\boxed{\text{20 goats}}$, and $\boxed{\text{25 sheep}}$.

$$
8x - 5x = 45
$$

$$
3x = 45
$$
$$
x = \frac{45}{3} = 15
$$

First quantity: $5x = 5 \times 15 = 75$  
Second quantity: $8x = 8 \times 15 = 120$  
Reasoning: By representing the quantities as multiples of a common factor $x$, the difference between them can be expressed in terms of $x$. Solving for $x$ allows us to find the actual values of the quantities while maintaining their given ratio.  
The two quantities are $\boxed{\text{75}}$ and $\boxed{\text{120}}$.