Due to some technical problems, an aeroplane started late by 1 hour from its starting point. The pilot decided to increase the speed of the aeroplane by 100 km/hr from its usual speed, to cover a journey of 1200 km in time. Find the usual speed of the aeroplane.

Hint: Assume the usual speed and time of the journey to be some variables. Use the given conditions and apply the formula ${\text{distance}} = {\text{speed}} \times {\text{time}}$ to form equations for usual journey and journey with late start and increased speed. Solve these two equations to get the usual speed of the aeroplane.

Complete step-by-step solution:
According to the question, the aeroplane started late by 1 hour due to some technical issues. And also, the pilot increases the speed by 100 km/hr.
The usual speed of the train is $s{\text{ km/hr}}$ and the time taken by the aeroplane to cover the journey is $t{\text{ hrs}}$.
Now, the distance to cover for the journey is given as 1200 km. So applying the formula ${\text{distance}} = {\text{speed}} \times {\text{time}}$, we’ll get:
$ \Rightarrow s \times t = 1200{\text{ }}.....{\text{(1)}}$
Now, since the pilot started 1 hour late, the time taken for the journey is \[\left( {t - 1} \right){\text{ hrs}}\] and he also increases the speed by 100 km/hr. So the increased speed of the aeroplane is $\left( {s + 100} \right){\text{ km/hr}}$. Again applying the same formula, we’ll get:
$ \Rightarrow \left( {s + 100} \right) \times \left( {t - 1} \right) = 1200$
Solving it further, we’ll get:
$ \Rightarrow s \times t - s + 100t - 100 = 1200$
Substituting $s \times t = 1200$ from equation (1), we’ll get:
$
   \Rightarrow 1200 - s + 100t - 100 = 1200 \\
   \Rightarrow 100t - s = 100
 $
Again putting $s = \dfrac{{1200}}{t}$ from equation (1), we’ll get:
$
   \Rightarrow 100t - \dfrac{{1200}}{t} = 100 \\
   \Rightarrow t - \dfrac{{12}}{t} = 1 \\
   \Rightarrow {t^2} - t - 12 = 0
 $
Solving quadratic equation using factorisation, we’ll get:
$
   \Rightarrow {t^2} - 4t + 3t - 12 = 0 \\
   \Rightarrow t\left( {t - 4} \right) + 3\left( {t - 4} \right) = 0 \\
   \Rightarrow \left( {t + 3} \right)\left( {t - 4} \right) = 0
 $
Equating both the factors to zero, we’ll get:
$
   \Rightarrow \left( {t + 3} \right) = 0{\text{ or }}\left( {t - 4} \right) = 0 \\
   \Rightarrow t = - 3{\text{ or }}t = 4
 $
Since time can never be negative, $t = 4$ is the only valid solution. Thus the usual time taken by the aeroplane for the journey is 4 hours.
Putting $t = 4$ in equation (1), we’ll get:
$
   \Rightarrow s \times 4 = 1200 \\
   \Rightarrow s = \dfrac{{1200}}{4} = 300
 $
Thus the usual speed of the aeroplane is 300 km/hr.

Note: If we are facing any difficulty solving a quadratic equation using factorization method, we can also use a direct formula to find its roots. Let the quadratic equation be:
$ \Rightarrow y = a{x^2} + bx + c$
The formula to determine its roots is:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$