P镇和Q镇相距 40 km。一辆汽车以 x km/h 的平均速度从P镇驶往Q镇。在返程中，该汽车随后以 (x+8) km/h 的平均速度从Q镇返回P镇，返程所花的时间减少了 12 分钟。

1. Express the time taken for each journey in hours.

Distance between the towns is $40$ km.

Outward journey time:
$\frac{40}{x}\text{ h}$

Return journey time:
$\frac{40}{x+8}\text{ h}$

2. Convert the time difference into hours and form the equation.

The return journey took $12$ minutes less.

$12\text{ min}=\frac{12}{60}\text{ h}=\frac{1}{5}\text{ h}$

Hence,
$\frac{40}{x}-\frac{40}{x+8}=\frac{1}{5}$

3. Clear the denominators.

Multiply throughout by $5x(x+8)$:

$5x(x+8)\left(\frac{40}{x}-\frac{40}{x+8}\right)=5x(x+8)\left(\frac{1}{5}\right)$

$200(x+8)-200x=x(x+8)$

4. Simplify to the required quadratic equation.

$200x+1600-200x=x^2+8x$

$1600=x^2+8x$

$x^2+8x-1600=0$

1. Identify the coefficients and substitute into the quadratic formula.

For $x^{2}+8x-1600=0$,

$a=1,\quad b=8,\quad c=-1600$

Using

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

$x=\frac{-8\pm\sqrt{8^{2}-4(1)(-1600)}}{2(1)}.$

2. Simplify the discriminant and expression for the roots.

$x=\frac{-8\pm\sqrt{64+6400}}{2}$

$x=\frac{-8\pm\sqrt{6464}}{2}$

3. Evaluate both roots and round to two decimal places.

$x=\frac{-8+\sqrt{6464}}{2}=36.1995\ldots$

$x=\frac{-8-\sqrt{6464}}{2}=-44.1995\ldots$

Correct to two decimal places,

$x=36.20\quad \text{or}\quad x=-44.20.$

1. Choose the physically valid value of $x$ as the car's speed from Town P to Town Q.

From part (ii),

$x=36.20\text{ or }-44.20.$

Since speed cannot be negative,

$x=36.20\text{ km/h}.$

2. Find the time taken to travel from Town P to Town Q.

Distance from Town P to Town Q $=40$ km.

$\text{time}=\frac{40}{36.20}=1.10497\ldots\text{ h}$

Convert to hours and minutes:

$0.10497\ldots\times 60=6.298\ldots\text{ min}$

So the journey time is approximately

$1\text{ h }6\text{ min}.$

3. Add the journey time to the departure time.

The car left Town P at 0900.

$0900+1\text{ h }6\text{ min}=1006.$