Question

An aircraft pilot flies it a distance of $800km$ with some average speed. He could have saved $40minutes$ by increasing the average speed of the aircraft by $40km/hr$. Find the actual speed of the aircraft.

Answer 1

Step 1: Given data

Let the actual speed of the aircraft be $xkm/hr$.

Distance covered by the aircraft is $800km$.

Time taken to cover the distance will be $t=\frac{dis\tan ce}{speed}$.

$\Rightarrow t=\frac{800}{x}$

If speed of the aircraft is increased by $40km/hr$ ,

Time taken to cover the same distance will be,

$\Rightarrow T=\frac{800}{x+40}$

Step 2: Determining the quadratic equation

According to the question,

Pilot could have saved $40minutes$ by increasing the average speed of the aircraft.

$$\begin{gathered} \Rightarrow t-T=40minutes \\ \Rightarrow \frac{800}{x}-\frac{800}{x+40}=\frac{40}{60}hours \\ \Rightarrow \frac{800\left(x+40\right)-800x}{x\left(x+40\right)}=\frac{2}{3} \\ \Rightarrow \frac{800x+32000-800x}{x\left(x+40\right)}=\frac{2}{3} \end{gathered}$$

After simplification,

$$\begin{gathered} \Rightarrow 3\times 32000=2x\left(x+40\right) \\ \Rightarrow 2x^2+80x-96000=0 \\ \Rightarrow 2\left(x^2+40x-48000\right)=0 \\ \Rightarrow x^2+40x-48000=0 \end{gathered}$$

Step 3: Determining the speed of the aircraft

To solve the obtained equation by split the middle term method, we choose two numbers such as their sum is equal to $b$ and product is equal to $a\times c$.

$$\begin{gathered} \Rightarrow x^2+\left(240-200\right)x-48000=0 \\ \Rightarrow x^2+240x-200x-48000=0 \\ \Rightarrow x\left(x+240\right)-200\left(x+240\right)=0 \\ \Rightarrow \left(x+240\right)\left(x-200\right)=0 \end{gathered}$$

After simplification,

$x=-240,200$

Here, $x=-240$ is not considerable as speed cannot be negative.

Therefore, speed of the aircraft is $200km/hr$.