Car A travels x km for every litre of petrol, while car B travels (x + 5) km for every litre of petrol.

(i) Write down the number of litres used by car A and car B in covering a distance of 400 km.

(ii) If car A used 4 litres of petrol more than car B in covering 400 km, write an equation in x and solve it to determine the number of litres of petrol used by car B for the journey.

(i) Given,

Car A travels x km for every litre of petrol.

Thus, car A travels 1 km in $\dfrac{1}{x}$ litres of petrol.

Thus, car A travels 400 km in $\dfrac{400}{x}$ litres of petrol.

Car B travels x + 5 km for every litre of petrol.

Thus, car B travels 1 km in $\dfrac{1}{x + 5}$ litres of petrol.

Thus, car B travels 400 km in $\dfrac{400}{x + 5}$ litres of petrol

Hence, number of litres used by car A and car B in covering a distance of 400 km = $\dfrac{400}{x}$ and $\dfrac{400}{x + 5}$ respectively.

(ii) Given,

Car A uses 4 lires more to cover 400 km.

$\Rightarrow \dfrac{400}{x} - \dfrac{400}{x + 5} = 4 \\[1em] \Rightarrow \dfrac{400(x + 5) - 400x}{x(x + 5)} = 4 \\[1em] \Rightarrow \dfrac{400x + 2000 - 400x}{x^2 + 5x} = 4 \\[1em] \Rightarrow \dfrac{2000}{x^2 + 5x} = 4 \\[1em] \Rightarrow 2000 = 4(x^2 + 5x) \\[1em] \Rightarrow 2000 = 4x^2 + 20x \\[1em] \Rightarrow 0 = 4x^2 + 20x - 2000 \\[1em] \Rightarrow 4(x^2 + 5x - 500) = 0 \\[1em] \Rightarrow x^2 + 5x - 500 = 0 \\[1em] \Rightarrow x^2 + 25x - 20x - 500 = 0 \\[1em] \Rightarrow x(x + 25) - 20(x + 25) = 0 \\[1em] \Rightarrow (x - 20)(x + 25) = 0 \\[1em] \Rightarrow (x - 20) = 0 \text{ or }(x + 25) = 0 \text{….[Using zero-product rule]} \\[1em] \Rightarrow x = 20 \text{ or } x = -25.$

Since, distance cannot be negative.

Thus, x = 20.

Petrol used by car B = $\dfrac{400}{x + 5} = \dfrac{400}{20 + 5} = \dfrac{400}{25}$ = 16 litres.

Hence, obtained equation is $\dfrac{400}{x} - \dfrac{400}{x + 5} = 4$ and the petrol used by car B = 16 litres.