阿兰（Alan）行驶了 $x$ 千米，他的汽车消耗了 7.5 升燃油。
写出一个关于 $x$ 的表达式，表示他的汽车每 100 千米的耗油量（单位：升）。 答案 ... 升每 100 千米 本（Ben）行驶了 $(x + 30)$ 千米，他的汽车消耗了 8 升燃油。
写出一个关于 $x$ 的表达式，表示他的汽车每 100 千米的耗油量（单位：升）。 答案 ... 升每 100 千米 阿兰的汽车每 100 千米的耗油量比本的汽车每 100 千米的耗油量多 1.5 升。
根据这一信息列出方程，并证明其可以化简为 $3x^2 + 190x - 45000 = 0$。 答案 (d) 解方程 $3x^2 + 190x - 45000 = 0$，两个答案均保留两位小数。 答案 x = ... 或 ... (e) 每升燃油的价格为 $2.42。
计算如果阿兰行驶 120 千米，他需要支付的汽油费用。 答案 $ ...

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.