# A plane left \[30\] minutes late than its scheduled time and in order to reach the destination \[1500km\] away in time, it had to increase its speed by \[100km/h\] from the usual speed. Find its usual speed.

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Hint: Assume the usual speed of the plane to be \[xkm/h\]. Use the fact that speed of plane is the ratio of distance travelled with time taken to travel the given distance, form a linear equation in one variable using the given data and solve the equations to get the usual speed of the plane.

Complete step-by-step answer:

We have to find the usual speed of the plane given the data related to the distance it travels over time.

Let’s assume that the usual speed of the plane is \[xkm/h\].

We know that the speed of a plane is the ratio of distance travelled with time taken to travel the given distance.

The plane travels at a speed of \[xkm/h\] to cover the distance of \[1500km\]. So the time taken by plane to cover the distance at a given speed is the ratio of distance covered by the plane to the speed of the plane.

Thus, we have time taken by plane \[=\dfrac{1500}{x}h\].

We know that if the plane left \[30\] minutes late, it covered the same distance of \[1500km\] by increasing its speed by \[100km/h\].

As we know that \[1hr=60\min \], dividing the equation by \[2\] on both sides, we have \[30\min =\dfrac{1}{2}hr\].

Thus, the delayed time of plane \[=\dfrac{1500}{x}-\dfrac{1}{2}h\] and new speed of plane \[=x+100km/h\].

Thus, we have \[x+100=\dfrac{1500}{\dfrac{1500}{x}-\dfrac{1}{2}}\].

Further simplifying the equation, we have \[\dfrac{1500}{x}-\dfrac{1}{2}=\dfrac{1500}{x+100}\].

Rearranging the terms, we get \[\dfrac{1500}{x}-\dfrac{1500}{x+100}=\dfrac{1}{2}\].

Thus, we have \[\dfrac{1500\left( x+100 \right)-1500x}{x\left( x+100 \right)}=\dfrac{1}{2}\].

\[\begin{align}

& \Rightarrow \dfrac{1500\times 100}{x\left( x+100 \right)}=\dfrac{1}{2} \\

& \Rightarrow x\left( x+100 \right)=2\left( 1500\times 100 \right) \\

& \Rightarrow {{x}^{2}}+100x-300000=0 \\

& \Rightarrow {{x}^{2}}+600x-500x-300000=0 \\

& \Rightarrow x\left( x+600 \right)-500\left( x+600 \right)=0 \\

& \Rightarrow \left( x+600 \right)\left( x-500 \right)=0 \\

& \Rightarrow x=500,-600 \\

\end{align}\]

As the speed can’t be a negative quantity, we have \[x=500km/h\] as the usual speed of the plane.

Hence, the usual speed of the plane is \[500km/h\].

Note: We can also solve this question by forming linear equations in two variables taking \[x\] as the speed of the plane and \[y\] as the time taken by the plane to cover the distance, and then solve those equations to find the speed of the plane.

Complete step-by-step answer:

We have to find the usual speed of the plane given the data related to the distance it travels over time.

Let’s assume that the usual speed of the plane is \[xkm/h\].

We know that the speed of a plane is the ratio of distance travelled with time taken to travel the given distance.

The plane travels at a speed of \[xkm/h\] to cover the distance of \[1500km\]. So the time taken by plane to cover the distance at a given speed is the ratio of distance covered by the plane to the speed of the plane.

Thus, we have time taken by plane \[=\dfrac{1500}{x}h\].

We know that if the plane left \[30\] minutes late, it covered the same distance of \[1500km\] by increasing its speed by \[100km/h\].

As we know that \[1hr=60\min \], dividing the equation by \[2\] on both sides, we have \[30\min =\dfrac{1}{2}hr\].

Thus, the delayed time of plane \[=\dfrac{1500}{x}-\dfrac{1}{2}h\] and new speed of plane \[=x+100km/h\].

Thus, we have \[x+100=\dfrac{1500}{\dfrac{1500}{x}-\dfrac{1}{2}}\].

Further simplifying the equation, we have \[\dfrac{1500}{x}-\dfrac{1}{2}=\dfrac{1500}{x+100}\].

Rearranging the terms, we get \[\dfrac{1500}{x}-\dfrac{1500}{x+100}=\dfrac{1}{2}\].

Thus, we have \[\dfrac{1500\left( x+100 \right)-1500x}{x\left( x+100 \right)}=\dfrac{1}{2}\].

\[\begin{align}

& \Rightarrow \dfrac{1500\times 100}{x\left( x+100 \right)}=\dfrac{1}{2} \\

& \Rightarrow x\left( x+100 \right)=2\left( 1500\times 100 \right) \\

& \Rightarrow {{x}^{2}}+100x-300000=0 \\

& \Rightarrow {{x}^{2}}+600x-500x-300000=0 \\

& \Rightarrow x\left( x+600 \right)-500\left( x+600 \right)=0 \\

& \Rightarrow \left( x+600 \right)\left( x-500 \right)=0 \\

& \Rightarrow x=500,-600 \\

\end{align}\]

As the speed can’t be a negative quantity, we have \[x=500km/h\] as the usual speed of the plane.

Hence, the usual speed of the plane is \[500km/h\].

Note: We can also solve this question by forming linear equations in two variables taking \[x\] as the speed of the plane and \[y\] as the time taken by the plane to cover the distance, and then solve those equations to find the speed of the plane.

Last updated date: 26th Sep 2023

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Total views: 363k

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