Question

# An aeroplane left 50 minutes later than its schedule time, and in order to reach thedestination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed.Itâ€™s is usual speed in km/hr is?

Hint: Assume the usual speed of the aeroplane as â€˜sâ€™ in km/hr. Since it has to cover a distance of 1250 km, find the usual time by the formula time = distance divided by speed. To cover 50 minutes,
it had to increase its speed by 250 km/hr i.e. itâ€™s speed is now â€˜s+250â€™. Find the time corresponding
to this speed and use the information given in the question to form an equation in â€˜sâ€™.

Let us assume the usual speed of the aeroplane as s km/hr. Since the aeroplane has to cover 1250
km, the usual time taken is $\dfrac{1250}{s}................\left( 1 \right)$.
To reach in time, it had increased itâ€™s speed by 250 km/hr. So, itâ€™s speed now becomes s+250. The
time taken by the aeroplane to cover 1250 km by this speed is $\dfrac{1250}{s+250}...........\left( 2 \right)$.
The speed of the aeroplane was increased in order to cover the time of 50 minutes. So the time we
have calculated in equation $\left( 1 \right)$ is 50 minutes greater than the time in equation $\left( 2 \right)$. Since we have calculated the time in hours in equation $\left( 1 \right)$ and equation $\left( 2 \right)$, we will have to convert 50 minutes in hours.
We know that,
60 minutes = 1 hour
$\Rightarrow 1$ min = $\dfrac{1}{60}$ hour
$\Rightarrow 50$ min = $\dfrac{50}{60}$ hour
$\Rightarrow 50$ min = $\dfrac{5}{6}$ hour
So the time we have calculated in equation $\left( 1 \right)$ is $\dfrac{5}{6}$ hour greater than the
time in equation $\left( 2 \right)$.
\begin{align} & \dfrac{1250}{s}=\dfrac{1250}{s+250}+\dfrac{5}{6} \\ & \Rightarrow \dfrac{1250}{s}-\dfrac{1250}{s+250}=\dfrac{5}{6} \\ & \Rightarrow 1250\left( \dfrac{1}{s}-\dfrac{1}{s+250} \right)=\dfrac{5}{6} \\ & \Rightarrow 250\left( \dfrac{s+250-s}{s\left( s+250 \right)} \right)=\dfrac{1}{6} \\ & \Rightarrow s\left( s+250 \right)=250\times 250\times 6 \\ & \Rightarrow {{s}^{2}}+250s-375000=0...............\left( 3 \right) \\ \end{align}
We get a quadratic equation which can be solved by using the quadratic formula.
For a quadratic equation $a{{x}^{2}}+bx+c=0$, the roots can be found out by the formula,
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Using quadratic formula to find roots of the quadratic equation $\left( 3 \right)$, we get,
\begin{align} & s=\dfrac{-250\pm \sqrt{{{250}^{2}}-4.1.\left( -375000 \right)}}{2.1} \\ & \Rightarrow s=\dfrac{-250\pm \sqrt{{{250}^{2}}-4.1.\left( -375000 \right)}}{2.1} \\ & \Rightarrow s=\dfrac{-250\pm \sqrt{62500+1500000}}{2} \\ & \Rightarrow s=\dfrac{-250\pm \sqrt{1562500}}{2} \\ & \Rightarrow s=\dfrac{-250\pm 1250}{2} \\ & \Rightarrow s=500,s=-750 \\ \end{align}
Since the speed cannot be negative, s=500km/hr.

Note: There is a possibility that one may not convert the time 50 minutes to hour. But since the
distance is given in km and the speed is given in km/hr in the question, it is necessary to convert the
time from minutes to hours.