# An aeroplane left 50 minutes later than its schedule time, and in order to reach the

destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed.

It’s is usual speed in km/hr is?

Last updated date: 17th Mar 2023

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Answer

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Hint: Assume the usual speed of the aeroplane as ‘s’ 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. it’s speed is now ‘s+250’. Find the time corresponding

to this speed and use the information given in the question to form an equation in ‘s’.

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 it’s speed by 250 km/hr. So, it’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.

it had to increase its speed by 250 km/hr i.e. it’s speed is now ‘s+250’. Find the time corresponding

to this speed and use the information given in the question to form an equation in ‘s’.

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 it’s speed by 250 km/hr. So, it’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.

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