proportion, giving a renson: 242 Crtansio the values of a min b A car travels a distance of 120 km and consumes 10 litras of petrol. \begin{tabular}{l|l} 2.51 & How far can it traval on 15 litres of potrol? \\ 2.52 & \( \begin{array}{l}\text { What is the rate at which the car consumes petral in tiuras } \\ \text { por } \mathrm{km} \text { ? }\end{array} \) \end{tabular} An employee al in courier business finds that if each parcel weighs a, he can easity carry 28 parcels at a lima in a conlainor. How many eils can ine carry if each parcol woighs 4009 ? Marcollo invesis R21 500 at ABC Bank, which offors him on intornet rate \( .4 \% \) per year, compounded annually.

**2.51 How far can it travel on 15 litres of petrol?** The car travels 120 km using 10 litres. Since the distance is directly proportional to the amount of petrol, we can write \[ \frac{120\text{ km}}{10\text{ L}} = \frac{x\text{ km}}{15\text{ L}}. \] Solve for \( x \): \[ x = 120 \times \frac{15}{10} = 120 \times 1.5 = 180\text{ km}. \] --- **2.52 What is the rate at which the car consumes petrol in litres per km?** The car uses 10 litres to travel 120 km. Thus, the consumption rate \( r \) is \[ r = \frac{10\text{ L}}{120\text{ km}} = \frac{1}{12}\text{ L/km}. \] --- **Courier Business Problem** An employee finds that when each parcel weighs \( a \) (in consistent units), he can carry 28 parcels in a container. This tells us the maximum weight he can carry is \[ \text{Total capacity} = 28a. \] If instead each parcel weighs 4009 (in the same units as \( a \)), then the number of parcels \( n \) that he can carry is given by \[ n \times 4009 = 28a. \] Solving for \( n \), \[ n = \frac{28a}{4009}. \] This is the number of parcels he may carry if each weighs 4009. *Reasoning:* The total weight remains the same; hence, increasing the weight per parcel reduces the number of parcels by the same factor. --- **Bank Investment Problem** Marcollo invests R21,500 at an annual interest rate of \( 0.4\% \) compounded annually. For one year of investment, the amount \( A \) after one year is calculated by \[ A = P(1 + r), \] where - \( P = 21500 \) (the principal), - \( r = 0.004 \) (since \( 0.4\% = 0.004 \)). Thus, \[ A = 21500 \times (1 + 0.004) = 21500 \times 1.004 = 21586. \] So, after one year, the amount in the account is R21,586. *Reasoning:* The interest for one year is \( 21500 \times 0.004 = 86 \), and adding this to the principal yields the final amount.