Question

In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by 100 km/h and time increased by 30 minutes. Find the original duration of the flight.

Answer 1

Let the speed be x
Now, time taken at the normal speed is given by
$\frac{2800}{x}\mathrm{hr}$
As the speed is reduced by 100 km/h
Hence, time taken after the reduction in speed is given by
$\frac{2800}{x-100}\mathrm{hr}$
A.T.Q
$$\begin{gathered} \frac{2800}{x-100}=\frac{2800}{x}+\frac{30}{60} \\ \Rightarrow \frac{2800}{x-100}-\frac{2800}{x}=\frac{1}{2} \\ \Rightarrow \frac{2800x-2800x+280000}{x\left(x-100\right)}=\frac{1}{2} \end{gathered}$$
$$\begin{gathered} \Rightarrow x\left(x-100\right)=560000 \\ \Rightarrow x^2-100x-560000=0 \\ \Rightarrow x^2-800x+700x-560000=0 \\ \Rightarrow x\left(x-800\right)+700\left(x-800\right)=0 \\ \Rightarrow \left(x-800\right)\left(x+700\right)=0 \\ \Rightarrow x=800\mathrm{km}/\mathrm{h}\left[\because x\ne -700\right] \end{gathered}$$
Thus, the original duration of the flight is given by
$\frac{2800}{800}=\frac{7}{2}=3\frac{1}{2}\mathrm{or}3\mathrm{hours}30\mathrm{minutes}$