Alan drove a distance of $x$ km and his car used 7.5 litres of fuel.
Write down an expression, in terms of $x$, for the rate of fuel used by his car in litres per 100 km. Answer ... litres per 100 km Ben drove a distance of $(x + 30)$ km and his car used 8 litres of fuel.
Write down an expression, in terms of $x$, for the rate of fuel used by his car in litres per 100 km. Answer ... litres per 100 km The rate of fuel used by Alan's car is 1.5 litres per 100 kilometre more than the rate of fuel used by Ben's car.
Form an equation to represent this information and show that it can be simplified to $3x^2 + 190x - 45000 = 0$. Answer (d) Solve the equation $3x^2 + 190x - 45000 = 0$, giving both answers correct to two decimal places. Answer x = ... or ... (e) Each litre of fuel costs $2.42.
Calculate the amount Alan paid for the petrol if he drove 120 km. Answer $ ...

1. Convert Alan's fuel consumption to a rate per 100 km.

Alan uses $7.5$ litres for $x$ km.

Fuel used per $100$ km
$=\frac{7.5}{x}\times 100$
$=\frac{750}{x}$

1. Convert Ben's fuel consumption to a rate per 100 km.

Ben uses $8$ litres for $(x+30)$ km.

Rate in litres per 100 km:
$\frac{8}{x+30}\times 100=\frac{800}{x+30}$

1. Form the equation from the comparison of the two fuel rates.

From parts (a) and (b),

Alan's rate $=\dfrac{750}{x}$ and Ben's rate $=\dfrac{800}{x+30}$.

Since Alan's rate is $1.5$ litres per 100 km more than Ben's rate,

$\dfrac{750}{x}=\dfrac{800}{x+30}+1.5$

2. Clear the denominators by multiplying throughout by $x(x+30)$.

$\dfrac{750}{x}=\dfrac{800}{x+30}+1.5$

Multiply throughout by $x(x+30)$:

$750(x+30)=800x+1.5x(x+30)$

3. Expand and rearrange to the required quadratic equation.

$750x+22500=800x+1.5x^2+45x$

$750x+22500=1.5x^2+845x$

$0=1.5x^2+95x-22500$

Multiply by $2$:

$0=3x^2+190x-45000$

Hence,

$3x^2+190x-45000=0$

1. Identify the coefficients and apply the quadratic formula.

For $3x^2+190x-45000=0$,

$a=3$, $b=190$, $c=-45000$.

Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$,

$x=\dfrac{-190\pm\sqrt{190^2-4(3)(-45000)}}{2(3)}$

2. Evaluate the discriminant and simplify the expression for the roots.

$x=\dfrac{-190\pm\sqrt{36100+540000}}{6}$

$x=\dfrac{-190\pm\sqrt{576100}}{6}$

3. Find both solutions and round to two decimal places.

$x=\dfrac{-190+\sqrt{576100}}{6}=94.835\ldots$

$x=\dfrac{-190-\sqrt{576100}}{6}=-158.168\ldots$

Correct to two decimal places,

$x=94.84 \text{ or } x=-158.17$

1. Select the valid distance for Alan from the solutions in part (d).

From part (d), $x=94.84$ or $x=-158.17$.

Since distance cannot be negative, use $x=94.84$.

2. Find the amount of fuel Alan used for 120 km by direct proportion.

Alan used $7.5$ litres for $94.84$ km.

Fuel used for $120$ km:
$\frac{7.5}{94.84}\times 120=9.489\ldots \text{ litres}$

3. Multiply the number of litres by the cost per litre and round to cents.

Cost:
$9.489\ldots \times 2.42=22.965\ldots$

So the amount paid is $\char36{}22.97$, correct to the nearest cent.