Question

# 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.

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.

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.